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SSC GD Constable 10 February 2025 · Shift 3 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 73 of 210

10 February 2025 · Shift 3

Elementary Mathematics

20 questions ~20 min readFree

20 questions

Nitin has ₹1224 with him. He divided it amongst his sons Pravin and Rishi and asked them to invest it at 4% rate of interest compounded annually. It was seen that Pravin and Rishi got same amount after 14 and 15 years respectively. How much (in ₹) did Nitin give to Rishi?

A.624
B.600
C.724
D.450

Solution

Compound interest, equal amounts Let Pravin get P and Rishi get R, so P+R=1224. The two amounts are equal, but Rishi's money grows for one year longer:

P {(1 + \frac{4}{100})}^{14} = R {(1 + \frac{4}{100})}^{15}

Cancelling

{(\frac{26}{25})}^{14}

from both sides:

P = R \times \frac{26}{25}

So P:R=26:25, that is 26+25=51 equal parts in all.

1 part

= \frac{1224}{51} = 24

, so one part is ₹24. Rishi's share

= 25 \times 24 =

₹600 So the correct option is (b).

Evaluate:

16 + 12 \div 3 - 2 \times 4

A.15
B.11
C.14
D.12

Solution

BODMAS simplification BODMAS: do division and multiplication first, then addition and subtraction from the left.

16 + 12 \div 3 - 2 \times 4

=16+4-8 =20-8=12 So the correct option is (d).

400 chickooes were bought at ₹1250 per hundred and were sold at a profit of ₹800. Find the selling price (in ₹) per dozen of chickooes.

A.174
B.189
C.184
D.164

Solution

Profit and unit selling price 400 chickoos are 4 hundreds, and one hundred costs ₹1250. Cost price

= 4 \times 1250 =

₹5,000 Selling price of all 400 =5000+800= ₹5,800 Selling price of one chickoo

= \frac{5800}{400} =

₹14.50 Selling price per dozen

= 12 \times 14.50 =

₹174 So the correct option is (a).

A shopkeeper bought an article for ₹500. At what price (in ₹) should he sell the article to make 30% profit ?

A.650
B.350
C.638
D.662

Solution

Selling price from profit percent Profit percent is always reckoned on the cost price:

S P = C P \times (1 + \frac{profit %}{100})

S P = 500 \times (1 + \frac{30}{100})

S P = 500 \times \frac{130}{100}

SP= ₹650 So the correct option is (a).

Mohan invests a sum of ₹5400 and Varun invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 6 years, Varun gets ₹480 more interest than Mohan, then find the rate of interest per annum (in percentage).

A.4
B.2.5
C.3
D.2

Solution

Simple interest, equal rate and time

S I = \frac{P \times R \times T}{100}

, and here R and T are the same for both men. So the extra interest comes only from the extra principal. Extra principal =9400-5400= ₹4,000

\frac{4000 \times R \times 6}{100} = 480

240R=480

R = 2 %

per annum So the correct option is (d).

A man goes to Jaipur from Surat at a speed of

3 km / hr

and returns to Surat at speed of 15 km/hr, through same route. What is his average speed (in km/hr) of the entire journey ?

Solution

Average speed, equal distances The two legs cover the same distance, so use the harmonic mean

= \frac{2 x y}{x + y}

= \frac{2 \times 3 \times 15}{3 + 15}

= \frac{90}{18}

=5 km/hr The ordinary average

\frac{3 + 15}{2} = 9

is the trap: far more time is spent on the slow leg. So the correct option is (a).

The smallest natural number which is divisible by

11, 6, 55

and 9 is:

A.921
B.990
C.1054
D.1032

Solution

LCM by prime factorisation The smallest number divisible by all of them is their LCM. 11=11,

6 = 2 \times 3

55 = 5 \times 11

9 = 3^{2}

Take the highest power of each prime: LCM

= 2 \times 3^{2} \times 5 \times 11

= 2 \times 9 \times 55 = 990

So the correct option is (b).

The population of a district is 384000, out of which 192000 are males. 69% of the population is literate. If 53% males are literate, then what percentage of females are literate?

A.84%
B.87%
C.85%
D.82%

Solution

Percentage, literacy split Females =384000-192000=192000, so the district is half male and half female. Total literates

= \frac{69}{100} \times 384000 = 264960

Literate males

= \frac{53}{100} \times 192000 = 101760

Literate females =264960-101760=163200

Required percentage

= \frac{163200}{192000} \times 100 = 85 %

So the correct option is (c).

A cyclist rides for 2 hours at 15

km / hr

and then for 3 hours at

20 km / hr

. What is the average speed (in km/hr) of the cyclist for the entire trip ?

A.18
B.16
C.20
D.19

Solution

Average speed, unequal times Average speed

= \frac{total distance}{total time}

- never the average of the two speeds. Total distance

= 2 \times 15 + 3 \times 20 = 30 + 60 = 90

km Total time =2+3=5 hours Average speed

= \frac{90}{5} = 18

km/hr So the correct option is (a).

What is the volume of a sphere of radius

3 \sqrt{2} cm

A. $73 \sqrt{2} π {cm}^{3}$
B. $71 \sqrt{2} π {cm}^{3}$
C. $70 \sqrt{2} π {cm}^{3}$
D. $72 \sqrt{2} π {cm}^{3}$

Solution

Volume of a sphere Volume of a sphere

= \frac{4}{3} π r^{3}

Here

r = 3 \sqrt{2}

cm, so

r^{3} = 3^{3} \times {(\sqrt{2})}^{3} = 27 \times 2 \sqrt{2} = 54 \sqrt{2}

V = \frac{4}{3} π \times 54 \sqrt{2}

V = 4 \times 18 \sqrt{2} π = 72 \sqrt{2} π {cm}^{3}

So the correct option is (d).

3, 6, 12

and

b

are in continued proportion, then what is the value of

b

A. $b = 2$
B. $b = 24$
C. $b = 4$
D. $b = 12$

Solution

Continued proportion In continued proportion every term bears the same ratio to the next:

\frac{3}{6} = \frac{6}{12} = \frac{12}{b}

The common ratio is

\frac{3}{6} = \frac{1}{2}

, i.e. each term is twice the one before it.

\frac{12}{b} = \frac{1}{2}

b=24 So the correct option is (b).

The average weight of Kiran, Mahesh and Ram is 46 kg . If the average weight of Kiran and Mahesh be 40 kg and that of Mahesh and Ram be 47 kg, then the weight of Mahesh (in kg) is:

A.56
B.51
C.36
D.46

Solution

Averages of overlapping groups Turn every average back into a total before combining them.

K + M + R = 3 \times 46 = 138

K + M = 2 \times 40 = 80

M + R = 2 \times 47 = 94

Adding the last two counts Mahesh twice: K+2M+R=80+94=174 Subtracting the first total: M=174-138 M=36 kg So the correct option is (c).

If at the same rate of interest, in 2 years, the simple interest is ₹40 and compound interest is ₹56, then what is the principal (in ₹) ?

A.25
B.18
C.29
D.20

Solution

Difference of CI and SI For 2 years the whole difference between CI and SI is the interest earned on the first year's interest. First year's interest

= \frac{40}{2} =

₹20 CI-SI=56-40= ₹16 So

20 \times \frac{R}{100} = 16

R = 80 %

per annum

P = \frac{S I \times 100}{R \times T} = \frac{40 \times 100}{80 \times 2}

P= ₹25 So the correct option is (a).

The speed of a stream is 6 km/h. A boat can go 56 km downstream and 39 km upstream in 7 hours. What is the speed (in km/h) of the boat in still water?

A.22
B.15
C.7
D.13

Solution

Boat and stream Let the speed of the boat in still water be x km/h. Downstream speed =x+6 km/h and upstream speed =x-6 km/h.

\frac{56}{x + 6} + \frac{39}{x - 6} = 7

56 (x - 6) + 39 (x + 6) = 7 (x^{2} - 36)

95 x - 102 = 7 x^{2} - 252

7 x^{2} - 95 x - 150 = 0

x = \frac{95 \pm \sqrt{9025 + 4200}}{14} = \frac{95 \pm 115}{14}

x=15 km/h, the negative root being rejected. So the correct option is (b).

Find the value of [(14 ÷ 2) × {14/7 + 40/3 × (7 − 4)}]

A.310
B.274
C.277
D.294

Solution

BODMAS with nested brackets Clear the innermost brackets first, then the curly bracket, then the square bracket.

(14 \div 2) = 7

and (7-4)=3 Inside the curly bracket:

\frac{40}{3} \times 3 = 40

\frac{14}{7} + 40 = 2 + 40 = 42

7 \times 42 = 294

So the correct option is (d).

Evaluate:

(- 9) - (- 60) \div (- 15) + (- 4) \times 8

A.- 47
B.- 48
C.-44
D.-45

Solution

BODMAS with negative numbers Division and multiplication come before addition and subtraction; a negative divided by a negative is positive.

(- 60) \div (- 15) = 4

and

(- 4) \times 8 = - 32

(-9)-4+(-32) =-13-32 =-45 So the correct option is (d).

A shopkeeper offers 3 schemes of discounts to his customers. Which of the following schemes will fetch the minimum discount percentage? A. Two successive discounts of 25% and 32%. B. Buy 8 get two free. C. Buy 8 and get 10.

A.B only
B.A only
C.Both B and C
D.C only

Solution

Comparing discount schemes (A) Successive discounts multiply: the customer pays

\frac{75}{100} \times \frac{68}{100} = \frac{51}{100}

of the marked price. Discount

= 100 - 51 = 49 %

(B) Buy 8 get 2 free: 10 articles are taken away but only 8 are paid for, so the discount is

\frac{2}{10} \times 100 = 20 %

. (C) Buy 8 and get 10 is exactly the same bargain - pay for 8, carry away 10 - so it too is

20 %

The smallest discount is

20 %

, and B and C give it equally. So the correct option is (c).

P and Q can complete a piece of work in 28 days and 35 days, respectively. P begins to do the work and they work alternatively one at a time for one day each. In how many days will the whole work be completed ?

A.31
B.27
C.18
D.21

Solution

Alternate-day work Take the total work as LCM(28,35)=140 units. P does

\frac{140}{28} = 5

units a day and Q does

\frac{140}{35} = 4

units a day. Working alternately from P, one 2-day cycle finishes 5+4=9 units.

140 = 9 \times 15 + 5

, so 15 full cycles (30 days) finish 135 units.

Day 31 is P's turn, and P's 5 units exactly clear the remaining 5 units. Total time =30+1=31 days So the correct option is (a).

The power consumption of an electrical device varies directly with the square of the voltage applied. If a device consumes 150 watts when the voltage is 120 volts, how much power (in watts) will it consume at 240 volts ?

A.550
B.650
C.500
D.600

Solution

Direct variation with a square

P \propto V^{2}

, so

\frac{P_{1}}{V 12} = \frac{P_{2}}{V 22}

\frac{150}{120^{2}} = \frac{P_{2}}{240^{2}}

P_{2} = 150 \times {(\frac{240}{120})}^{2}

P_{2} = 150 \times 2^{2} = 150 \times 4

P_{2} = 600

watts Doubling the voltage does not double the power - it multiplies it by four. So the correct option is (d).

If the price of petrol increases from ₹90/L to ₹110/L, by what percentage does a person have to decrease his consumption so that his expenditure on petrol remains the same (approximate to two decimals) ?

A.78.78%
B.25.25%
C.18.18%
D.81.81%

Solution

Price rise and consumption cut Expenditure = price × consumption, and it has to stay unchanged. Required reduction

% = \frac{rise in price}{new price} \times 100

= \frac{110 - 90}{110} \times 100

= \frac{20}{110} \times 100 = \frac{200}{11}

= 18.18 %

Dividing by the old price would give

22.22 %

- the cut has to be measured on the new price. So the correct option is (c).

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