SELINA Solution Class 9 Chapter 13 Pythagoras Theorem [Proof and simple applications with converse] Exercise 13B

Question 1

In the figure, given below, AD ⊥ BC.
Prove that: c² = a² + b² - 2ax.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABD and applying Pythagoras theorem we get,
AB² = AD²   + BD² 
c²  = h²  + ( a - x )²  
h²  = c²  - ( a - x )²                      ......(i)
First, we consider the ΔACD and applying Pythagoras theorem we get,
AC² = AD² + CD² 
b²  = h² + x²  
h²  = b² - x²                              ......(ii)

From (i) and (ii) we get,
c²  - ( a - x )² = b² - x²  
c²  - a² - x²  + 2ax = b² - x² 
c² = a² + b²  - 2ax
Hence Proved.

Question 2

In equilateral Δ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

Sol:

In equilateral Δ ABC, AD ⊥ BC.
Therefore, BC = x cm.

Area of equilateral ΔABC = ${\displaystyle \frac{\sqrt{3}}{4}\times {\text{side}}^2 =\frac{1}{2}\times \text{base}\times \text{height}}$

= ${\displaystyle \frac{\sqrt{3}}{4}\times x^2=\frac{1}{2}\times x\times \text{AD}}$

AD = ${\displaystyle \frac{\sqrt{3}}{2}x}$

Question 3

ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM² + BC² = AC² + BM².

Sol:

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABM and applying Pythagoras theorem we get,
AM² = AB  + BM² 
AB² = AM² - BM²               ......(i)
Now, we consider the ΔABC and applying Pythagoras theorem we get,
AC² = AB²  + BC² 
AB² = AC² - BC²                ......(ii)

From (i) and (ii) we get,
AM²  - BM²  = AC²  - BC² 
AM²  + BC² = AC² + BM²  
Hence Proved.

Question 4

M and N are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²(iv) 4 (PM² + RN²) = 5 PR²

Sol:

We draw, PM, MN, NR

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since M and N are the mid-points of the sides QR and PQ respectively, therefore, PN = NQ, QM = RM

(i) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM²= PQ² + MQ²
= ( PN + NQ )² + MQ²
= PN + NQ² + 2PN . NQ + MQ²   
= MN²+ PN² + 2PN.NQ  ...[From, ΔMNQ, MN² = NQ² + MQ²]  ......(i)

Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN² = NQ² + RQ² 
= NQ² + ( QM + RM )²
= NQ²  + QM²  + RM²  + 2QM .RM
= MN²  + RM²  + 2QM . RM   .......(ii)

Adding (i) and (ii) we get, 
PM²  + RN²  = MN²  + PN²  + 2PN.NQ + MN²  + RM²  + 2QM. RM
PM²  + RN²  = 2MN²  + PN²  + RM²  + 2PN.NQ + 2QM.RM 
PM²  + RN²  = 2MN²  + NQ²  + QM²  + 2(QN² ) + 2(QM² )
PM²  + RN²  = 2MN²  + MN²  + 2MN² 
PM ² + RN²  = 5MN²                 Hence Proved.

(ii) We consider the ΔPQM, and applying Pythagoras theorem we get,
PM² = PQ² + MQ²
4PM² = 4PQ² + 4MQ²             ...[ Multiply both sides by 4]
4PM² = 4PQ² + 4.( 1/2 QR )² ...[ MQ = ${\displaystyle \frac{1}{2}}$ QR ]
4PM² = 4PQ² + 4PQ + 4 . ${\displaystyle \frac{1}{2}}$ QR²
4PM² = 4PQ²  + QR² 
Hence Proved.

(iii) We consider the ΔRQN, and applying Pythagoras theorem we get,
RN² = NQ² + RQ²
4RN² = 4NQ² + 4QR²  ...[ Multiplying both sides by 4]
4RN² = 4QR² + 4 .(1/2 PQ)² ...[ NQ = ${\displaystyle \frac{1}{2}}$ PQ ]
4RN² = 4QR² + 4 .${\displaystyle \frac{1}{4}}$ PQ²
4RN² = PQ² + 4QR²
Hence Proved.

(iv) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM² = PQ² + MQ²
= ( PN + NQ )² + MQ²
= PN² + NQ² + 2PN.NQ + MQ²   
= MN² + PN² + 2PN.NQ  ...[ From, ΔMNQ, = MN² = NQ² + MQ² ]  ......(i)

Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN² + NQ² + RQ² 
= NQ² + ( QM + RM )²
= NQ²  + QM²  + RM²  + 2QM .RM
=MN²  + RM²  + 2QM . RM .......(ii)

Adding (i) and (ii) we get,
PM²  + RN²  = MN²  + PN²  + 2PN . NQ + MN²  + RM²  + 2QM. RM
PM²  + RN²  = 2MN²  + PN²  + RM²  + 2PN . NQ + 2QM . RM 
PM²  + RN²  = 2MN²  + NQ²  + QM²  + 2(QN² ) + 2(QM² )
PM²  + RN²  = 2MN²  + MN²  + 2MN² 
PM ² + RN²  = 5MN² 
4( PM² + RN² ) = 4.5. (NQ² + MQ²)
4( PM² + RN² ) = 4.5. ${\displaystyle [{\left(\frac{1}{2}PQ\right)}^2+{\left(\frac{1}{2}RQ\right)}^2] ....[\because NQ=\frac{1}{2}PQ,MQ=\frac{1}{2}QR]}$
4 ( PM² + RN² ) = 5PR²
Hence Proved.

 

Question 5

In triangle ABC, ∠B = 90^o and D is the mid-point of BC.
Prove that: AC² = AD² + 3CD².

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

In triangle ABC, ∠B = 90^o and D is the mid-point of BC. Join AD. Therefore, BD = DC

First, we consider the ΔADB, and applying Pythagoras theorem we get,
AD² = AB² + BD²
AB² = AD² - BD²                 ....(i)

Similarly, we get from rt. angle triangles ABC we get,
AC² = AB² + BC²
AB² = AC² - BC²                .....(ii)
From (i) and (ii)
AC² - BC² = AD² - BD² 
AC² = AD² - BD² + BC²
AC² = AD² - CD² + 4CD²   ....[ BD = CD = ${\displaystyle \frac{1}{2}}$ BC ]
AC² = AD² + 3CD² 
Hence proved.

Question 6

In a rectangle ABCD,
prove that: AC² + BD² = AB² + BC² + CD² + DA².

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since, ABCD is a rectangle angles A, B, C and D are rt. angles.

First, we consider the ΔACD, and applying Pythagoras theorem we get,
AC² = DA² + CD²                  ....(i)

Similarly, we get from rt. angle triangle BDC we get,
BD² = BC² + CD²
= BC² + AB²          ....[ In a rectangle, opposite sides are equal, ∴ CD = AB ] ...(ii)

Adding (i) and (ii)
AC² + BD² = AB² + BC² + CD² + DA²
Hence proved.

Question 7

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC² - AB² = BC² + CD² + DA²

Sol:

In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.

In ΔABC, using Pythagoras theorem,
AC² = AB² + BC²
⇒ AB² = AC² - BC²                  ....(i)

In ΔADC, using Pythagoras theorem,
AC² = AD² + DC²                   ....(ii)

LHS = 2AC² - AB²
= 2AC² - ( AC² - BC² )           .....[ From(i) ]
= 2AC² - AC² + BC²
= AC² + BC²
= AD² + DC² + BC²            ....[ From(ii) ]
= RHS

Question 8

O is any point inside a rectangle ABCD.
Prove that: OB² + OD² = OC² + OA².

Sol:

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:
OA² = AH² + OH² = AH² + AE²
OC² = CG² + OG² = EB² + HD²
OB² = EO² + BE² = AH² + BE²
OD² = HD² + OH² = HD² + AE²

Adding these equalities we get:
OA² + OC² = AH² + HD² + AE² + EB²
OB² + OD² = AH² + HD² + AE² + EB²

From which we prove that for any point within the rectangle there is the relation
OA² + OC² = OB² + OD²
Hence Proved.

Question 9

In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.

Prove that: AR² + BP² + CQ² = AQ² + CP² + BR²

Sol:

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the ΔARO and applying Pythagoras theorem we get,

AQ² = AR² + OR²
AR² = AO² - OR²            ....(i)
Similarly, from triangles, BPO, COQ, AOQ, CPO, and BRO we get the following results,
BP² = BO² - OP²          ....(ii)
CQ² = OC² - OQ²        ....(iii)
AQ² = AO² - OQ²        ....(iv)
CP² = OC² - OP²          ....(v)
BR² = OB² - OR²          ....(vi)

Adding (i), (ii) and (iii), we get 
AR² + BP² + CQ² = AO² - OR² + BO² - OP² + OC² - OQ²  ....(vii)

Adding (iv), (v) and (vi), we get,
AQ² + CP² + BR² = AO² - OR² + BO² - OP² + OC² - OQ²  ....(viii)

From (vii) and (viii), we get,
AR² + BP² + CQ² = AQ² + CP² + BR²
Hence proved.

Question 10

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA² + OC² = 2AD² - ${\displaystyle \frac{{\text{BD}}^2}{2}}$

SOl:

Diagonals of the rhombus are perpendicular to each other.

In quadrilateral ABCD, ∠AOD = ∠COD = 90°.
So, ΔAOD and ΔCOD are right-angled triangles.

In ΔAOD using Pythagoras theorem,
AD² = OA² + OD²
⇒ OA² = AD² - OD²                 ....(i)

In ΔCOD using Pythagoras theorem,
CD² = OC² + OD²
⇒ OC² = CD² - OD²               ....(ii)

LHS = OA² + OC²
= AD² - OD² + CD² - OD²      ...[ From(i) and (ii) ]
= AD² + CD²  - 2OD²

= AD² + AD² - 2${\displaystyle {\left(\frac{\text{BD}}{2}\right)}^2}$ ...[ AD = CD and OD = ${\displaystyle \frac{\text{BD}}{2}}$]

= 2AD² - ${\displaystyle \frac{{\left(\text{BD}\right)}^2}{2}}$

= RHS.

Question 11

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC² = 2BC. DC.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the ΔACD and applying Pythagoras theorem we get,
AC² = AD² + DC²
= ( AB² - DB² ) + ( DB + BC )²
= BC² - DB² + DB² + BC² + 2DB.BC    ...( Given, AB = BC )
= 2BC² + 2DB.BC
= 2BC( BC + DB )
= 2BC . DC
Hence proved.

Question 12

In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD² = AC² + BD.CD.

Sol:

In an isosceles triangle ABC; AB = AC and
D is the point on BC produced.
Construct AE perpendicular BC.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled ΔAED and applying Pythagoras theorem we get,
AD² = AE² + ED²
AD² = AE² + ( EC + CD )²             ....(i)[ ∵ ED = EC + CD ]

Similarly, in ΔAEC,
AC² = AE² + EC²
AE² = AC² - EC²                       ....(ii)
Putting AE² = AC² - EC² in (i), We get,
AD² = AC² - EC² + ( EC + CD )²
        = AC² + CD( CD + 2EC )
AD² = AC² + BD.CD                 .....[ ∵ 2EC + CD = BD ]     
Hence proved.

Question 13

In triangle ABC, angle A = 90^o, CA = AB and D is the point on AB produced.
Prove that DC² - BD² = 2AB.AD.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled  ΔACD and applying Pythagoras theorem we get,
CD² = AC² + AD²
CD² = AC² + ( AB + BD )²                    ....[ ∵ AD = AB + BD ]
CD² = AC² + AB² + BD² + 2AB.BD      ...(i)

Similarly, in ΔABC,
BC² = AC² + AB²
BC² = 2AB²                                        ...[ AB = AC ]
AB² = ${\displaystyle \frac{1}{2}}$BC²                                   ...(ii)

Putting, AB2 from (ii) in (i), We get,
CD² = AC² + ${\displaystyle \frac{1}{2}}$BC² + BD² + 2AB . BD

CD² - BD² = AB² + AB² + 2AB . ( AD - AB )

CD² - BD² = AB² + AB² + 2AB . AD - 2AB² 

CD² - BD² = 2AB . AD

DC² - BD² = 2AB . AD                       

Hence Proved.

Question 14

In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD² - CD² = 2CD × AD.

Sol:

In right-angled ΔADB,
AB² = AD² + BD²
⇒ AD² = AB² - BD²                                      .....(i)

AC = AD + DC
⇒ AC² = ( AD + DC )²
⇒  AC² = AD² + DC² + 2AD x DC
⇒ AC² = AB² - BD² + DC² + 2AD x DC      ...[ From(i) ]
⇒ AC² = AC² - BD² + DC² + 2AD x DC     ...[ AB = AC ]
⇒ BD² - DC² = 2AD x DC.

Question 15

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.

Prove that : 2AC² = 2AB² + BC²

Sol:

Here,
BD : DC = 1 : 3.
⇒ BD = ${\displaystyle \frac{1}{4}\text{BC}\text{and}CD=\frac{3}{4}}$BC

AC² = AD² + CD² and AB² = AD² + BD² 
Therefore,
AC² - AB² = CD² - BD²

= ${\displaystyle {\left(\frac{3}{4}\text{BC}\right)}^2-{\left(\frac{1}{4}\text{BC}\right)}^2}$

= ${\displaystyle \frac{9}{16}{\text{BC}}^2- \frac{1}{16}{\text{BC}}^2}$

= ${\displaystyle \frac{1}{2}{\text{BC}}^2}$

∴ 2AC² - 2AB² = BC²
2AC² = 2AB² + BC²
Hence proved.

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13B

 Exercise 13 B

Question 1

AB and CD bisect each other at K. Prove that AC = BD .

(IMAGE TO BE ADDED)

Sol: AK= BK

$\angle A K C=\angle B K D$ (Vertically opposite angle )

CK = DK 

By SAS $\triangle A K C, \cong \triangle B K D$

C.P.C.T $\quad A C=B D$ Hence proved 

Question 2

The sides BA and CA have been produced such that BA = AD and CA = AE prove that DE||BC 

(IMAGE TO BE ADDED)

Sol: $A B=A D$

$\angle B A C=\angle E A D$  (Vertically opposite angle )

by $S A S \quad \triangle A B C \cong \triangle A E D$

by C.P.CT $\angle B=\angle D \quad \because B C \| D E$  Hence proved .

Question 3

$O A=O B, O C=O D$, $\angle A O B=\angle C O D$ prove that AC = BD 

(IMAGE TO BE ADDED)

Sol: 

$\begin{aligned} O A &=O B \\ \angle A O B &=\angle C O D \\ \angle A O C+\angle C O B=\angle C O B+\angle B O D \\ \angle A O C &=\angle B O D \\ O C &=O D \end{aligned}$

By SAS $\triangle A O C \cong \triangle B O D$

by C.P.CT $A C=B D$ hence proved 

Question 4

If AB and MN bisect each other at O and AM and BN are $\perp$ on xy. Prove that $\triangle S$, OAM and OBN are congruent and hence prove that AM= BN 

(IMAGE TO BE ADDED)

Sol: $\angle A M O=\angle B N O=90^{\circ}$

$O B=O A$

$O M=O N$

by RHS $\triangle A M O \cong \triangle B N O$ 

by $C \cdot P \cdot c \cdot T \quad A M=B N$

Question 5

$\angle XYZ$ is bisected by YP . L is any point on YP and MLN is Perpendicular to YP. Prove that LM = LN 

(image to be added)

Sol: $\angle YLM=\angle YL N=90^{\circ}$

YL = YL (Common)

$\angle M YL=\angle N YL$ (Bisect by YP )

by ASA $\triangle MY L \cong \triangle N Y L$

By C.P.C.T LM = LN Hence proved 

Question 6

In the fig, PM =PN , PM $\perp A B$ AND $P N \perp A C$ Prove $\triangle A M P \cong \triangle A N P$

Sol: (image to be added)

$\angle A N P=\angle A M P=90^{\circ}$ 

$A P=A P \quad($ Comimon $)$

$M P=N P \quad$ (given)

By R.H.S $\triangle A M P \cong \triangle A N P$ Hence proved 

Question 7

In the figure , AD = BC and AD ||BC. Prove that AB =DC 

Sol: (IMAGE TO BE ADDED)

AD=BC 

$\angle C A D=\angle A C B$ [Alt. angles]

AC = AC 

By ASA $\triangle A C D \cong \triangle A B C$

by C.P.CT. $A B=D C$ Hence proved 

Question 8

In the given figure, triangles ABC and DCB are right angles at A and D respectively and AC = DB . Prove that $\triangle A B C \cong \triangle D C B$

 (IMAGE TO BE ADDED)

Sol: $\angle B A C=\angle B D C=90^{\circ}$

$B C=B C=$ Common

$A C=D B$

By RHS $\triangle A B C \cong \triangle D C B$ Hence proved

S.chand class 6 Mathematics Chapter 13 Exercise 13B

  Exercise 13B

Question 1

Find the value of each expression using the given values.

(i) $\frac{a b}{c}$ if $a=5, b=6$ and $c=10$

(ii) $\frac{x}{3}+8 y$, if $x=24$ and $y=2$

(iii) $2 x-y+3 z$, if $x=4, y=7$ and $z=2$

(iv) $\frac{p+6 q}{5 p-3 q}$, if $p=6$ and $q=2$

Question 2

If $x=3, y=2$ and $z=5$, find the value of

(i) $\frac{x^{2}}{9}+\frac{z^{2}}{25}$

(ii) $x^{2} y+x y^{2}$

(iii) $6(x+3 y)-(5 z-x)$

Question 3

What is the value of $3 x+4 y-6$, when $x=2 \frac{1}{3}$ and $y=4 \frac{3}{4}$ ?

Question 4

Evaluate $3 a+2 b+c$, when $a=1.2, b=3.7$ and $c=1.4 ?$

Multiple Choice Question (MCQ) 

Tick (✔) the correct option.

Question 5

The value of $b-a-8$ when $b=-3$ and $a=-11$ is

(a) $-22$

(b) 16

(c) 0

(d) $-16$

RS Aggarwal solution class 8 chapter 13 Time and Work Exercise 13B

Exercise 13B

Page-174

Q1 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 1:

Tick (✓) the correct answer:
A alone can do a piece of work in 10 days and B alone can do it in 15 days. In how many days will A and B together do the same work?
(a) 5 days
(b) 6 days
(c) 8 days
(d) 9 days

Answer 1:

(b) 6 days

​

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Question 2:

Tick (✓) the correct answer:
A man can do a piece of work in 5 days. He and his son working together can finish it in 3 days. In how many days can the son do it alone?
(a) $6\frac{1}{2}$ days
(b) 7 days
(c) $7\frac{1}{2}$ days
(d) 8 days

Answer 2:

(c) $7\frac{1}{2} days$

​

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Question 3:

Tick (✓) the correct answer:
A can do a job in 16 days and B can do the same job in 12 days. With the help of C, they can finish the job in 6 days only. Then, C alone can finish it in
(a) 34 days
(b) 22 days
(c) 36 dyas
(d) 48 days

Answer 3:

(d) 48 days

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Question 4:

Tick (✓) the correct answer:
To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?
(a) 30 days
(b) 35 days
(c) 40 days
(d) 45 days

Answer 4:

(a) 30 days

$$\begin{gathered} Let B\hspace{0.17em}take x days to complete the work. \\ Then A takes \left(x+\frac{50}{100}x\right)=1.5x \\ A\text{'}s 1 day\text{'}s work = \frac{1}{1.5x} = \frac{2}{3x} \\ B\text{'}s 1 day\text{'}s work = \frac{1}{x} \\ (A+B)\hspace{0.17em}takes 18 days to complete the work. \\ (A+B)\text{'}s 1 day\text{'}s net work = \frac{1}{18} \\ or \frac{1}{18}=\frac{2}{3x}+\frac{1}{x} \\ \Rightarrow \frac{1}{18}=\frac{5}{3x} \\ By cross-multiplication, we get: \\ x=30 days \\ \therefore B alone will take 30 days to complete the work. \end{gathered}$$

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Question 5:

Tick (✓) the correct answer:
A works twice as fast as B. If both of them can together finish a piece of work in 12 days, then B alone can do it in
(a) 24 days
(b) 27 days
(c) 36 days
(d) 48 days

Answer 5:

(c) 36 days

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Question 6:

Tick (✓) the correct answer:
A alone can finish a piece of work in 10 days which B alone can do in 15 days. If they work together and finish it, then out of total wages of Rs 3000, A will get
(a) Rs 1200
(b) Rs 1500
(c) Rs 1800
(d) Rs 2000

Answer 6:

(c) Rs. 1800

Since the wage distribution will follow the work distribution ratio, we have:

Work done by A in 1 day $=\frac{1}{10}$
Work done by B in 1 day $=\frac{1}{15}$

Net work done by (A+B) in 1 day $=\frac{1}{10}+\frac{1}{15}=\frac{5}{30}=\frac{1}{6}$

i.e., (A+B) will take 6 days to complete the work.

A's share of work in a day = $\frac{1}{10}÷\frac{1}{6}=\frac{1}{10}\times \frac{6}{1}=\frac{6}{10}=\frac{3}{5}$

∴ A's wage = $\frac{3}{5}\times 3000=\mathrm{Rs} 1800$

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Question 7:

Tick (✓) the correct answer:
The rates of working of A and B are in the ratio 3 : 4. The number of days taken by them to finish the work are in the ratio
(a) 3 : 4
(b) 9 : 16
(c) 4 : 3
(d) 16 : 9

Answer 7:

(c) 4:3

The number of days taken for working is the reciprocal of the rate of work.

Q8 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 8:

Tick (✓) the correct answer:
A and B together can do a piece of work in 12 days; B and C can do it in 20 days while C and A can do it in 15 days. A, B and C all working together can do it in
(a) 6 days
(b) 9 days
(c) 10 days
(d) $10\frac{1}{2}$ days

Answer 8:

(c) 10 days

Q9 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 9:

Tick (✓) the correct answer:
3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to do it?
(a) 6 days
(b) 5 days
(c) 4 days
(d) 3 days

Answer 9:

(c) 4 days

The work can be completed in 4 days.

Q10 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 10:

Tick (✓) the correct answer:
A can do a piece of work in 15 days. B is 50% more efficient than A. B can finish it in
(a) 10 days
(b) $7\frac{1}{2}$ days
(c) 12 days
(d) $10\frac{1}{2}$ days

Answer 10:

(a) 10 days

Page-175

Q11 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 11:

Tick (✓) the correct answer:
A does 20% less work than B. If A can finish a piece of work in $7\frac{1}{2}$ hours, then B can finish it in
(a) 5 hours
(b) $5\frac{1}{2}$ hours
(c) 6 hours
(d) $6\frac{1}{2}$ hours

Answer 11:

(c) 6 hours

Q12 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 12:

Tick (✓) the correct answer:
A can do a piece of work in 20 days which B alone can do in 12 days. B worked at it for 9 days. A can finish the remaining work in
(a) 3 days
(b) 5 days
(c) 7 days
(d) 11 days

Answer 12:

(b) 5 days

Q13 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 13:

Tick (✓) the correct answer:
A can do a piece of work in 25 days, which B alone can do in 20 days. A started the work and was joined by B after 10 days. The work lasted for
(a) $12\frac{1}{2}$ days
(b) 15 days
(c) $16\frac{2}{3}$ days
(d) 14 days

Answer 13:

(c) 

∴ Time taken if they work together=35÷9100=35×1009=203=623 days

Q14 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 14:

Tick (✓) the correct answer:
Two pipes can fill a tank in 20 minutes and 30 minutes respectively. If both the pipes are opened simultaneously, then the tank will be filled in
(a) 10 minutes
(b) 12 minutes
(c) 15 minutes
(d) 25 minutes

Answer 14:

(b) 12 minutes

Q15 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 15:

Tick (✓) the correct answer:
A tap can fill a cistern in 8 hours and another tap can empty the full cistern in 16 hours. If both the taps are open, the time taken to fill the cistern is
(a) $5\frac{1}{3}$ hours
(b) 10 hours
(c) 16 hours
(d) 20 hours

Answer 15:

(c) 16 hours

Q16 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 16:

Tick (✓) the correct answer:
A pump can fill a tank in 2 hours. Due to a leak in the tank it takes $2\frac{1}{3}$ hours to fill the tank. The leak can empty the full tank in
(a) $2\frac{1}{3}$ hours
(b) 7 hours
(c) 8 hours
(d) 14 hours

Answer 16:

(d) 14 hours

Q17 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 17:

Tick (✓) the correct answer:
Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be full?
(a) 7 hrs 15 min
(b) 7 hrs 30 min
(c) 7 hrs 45 min
(d) 8 hrs

Answer 17:

(b) 7 hours 30 minutes