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SSC GD Constable 25 February 2025 · Shift 2 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 203 of 210

25 February 2025 · Shift 2

Elementary Mathematics

20 questions ~20 min readFree

20 questions

If the average marks of three sections of Class IX, having 30, 45 and 50 students, respectively, is 80,75 and 90 , then the average marks of all the students of Class IX, is:

A.85.6
B.82.2
C.78.4
D.80.9

Solution

Weighted average Total marks of a section = number of students

\times

average marks. Section 1:

30 \times 80 = 2400

Section 2:

45 \times 75 = 3375

Section 3:

50 \times 90 = 4500

Combined average

= \frac{2400 + 3375 + 4500}{30 + 45 + 50}

= \frac{10275}{125} = 82.2

Hence the average marks of the whole class is 82.2, so the correct option is (b).

Compare the ratios 12 : 13, 7 : 8, 9 : 10 and 5 : 6. Which one is greater than 0.9 but smaller than 0.95?

A.9 : 10
B.7 : 8
C.12 : 13
D.5 : 6

Solution

Comparing ratios Write every ratio as a decimal and then compare.

\frac{12}{13} = 0.923

\frac{7}{8} = 0.875

\frac{9}{10} = 0.900

\frac{5}{6} = 0.833

Only 0.923 lies between 0.9 and 0.95; note that

\frac{9}{10}

is exactly 0.9, so it is not greater than 0.9. Hence the required ratio is 12 : 13, so the correct option is (c).

To finish a work in 8 days, 10 men are required, or 15 women are required. If '

x

' men and '

y

' women are required to finish the work in 12 days, then which of the following equations is correct ?

A. $3 x + 2 y = 20$
B. $2 x + 3 y = 30$
C. $2 x + 3 y = 20$
D. $3 x + 2 y = 30$

Solution

Time and work equation Work = efficiency

\times

time, and the same work is done in each case. Take the total work as 120 units (LCM of

10 \times 8

and

15 \times 8

). 1 man's one day work

= \frac{120}{10 \times 8} = \frac{120}{80} = \frac{3}{2}

units 1 woman's one day work

= \frac{120}{15 \times 8} = \frac{120}{120} = 1

unit

x men and y women finish the same work in 12 days:

12 (\frac{3}{2} x + y) = 120

\frac{3}{2} x + y = 10

Multiplying throughout by 2: 3x+2y=20 Hence the correct equation is 3x+2y=20, so the correct option is (a).

Ashok has ₹1612 with him. He divided it amongst his sons Raj and Varun and asked them to invest it at 8% rate of interest compounded annually. It was seen that Raj and Varun got same amount after 16 and 17 years respectively. How much (in ₹) did Ashok give to Raj ?

A.837
B.875
C.775
D.687

Solution

Compound interest division Amount at compound interest

= P {(1 + \frac{8}{100})}^{n} = P {(\frac{27}{25})}^{n}

. Let Raj get R and Varun get V, so R+V=1612. The two amounts are equal, so

R {(\frac{27}{25})}^{16} = V {(\frac{27}{25})}^{17}

\frac{R}{V} = {(\frac{27}{25})}^{17 - 16} = \frac{27}{25}

So R:V=27:25 and the total is 27+25=52 parts.

52 parts

= ₹ 1612 \Rightarrow 1

part

= \frac{1612}{52} = 31

Raj's share

= 27 \times 31 = 837

Hence Ashok gave Raj ₹837, so the correct option is (a).

Two trains are moving in opposite directions at speeds of

140 km / h

and 80

km / h

. The length of one train is 340 m . The time taken by them to cross each other is 11 seconds. The length (in m ) of the other train, correct to 2 decimal places, is:

A.330.83
B.331.39
C.332.22
D.330.66

Solution

Trains crossing each other The trains move in opposite directions, so the relative speed is the sum of their speeds. Relative speed =140+80=220 km/h 220 km/h

= 220 \times \frac{5}{18} = \frac{1100}{18}

m/s

While crossing each other they together cover the sum of the two lengths. Sum of lengths = relative speed

\times

time

= \frac{1100}{18} \times 11 = \frac{12100}{18} = 672.22

m Length of the other train =672.22-340=332.22 m Hence the other train is 332.22 m long, so the correct option is (c).

Evaluate:

(- 9) - (- 60) \div (- 12) + (- 4) \times 7

A.- 41
B.- 45
C.- 42
D.- 44

Solution

BODMAS with integers By BODMAS, division and multiplication are done first, then addition and subtraction.

(- 60) \div (- 12) = 5

(- 4) \times 7 = - 28

The expression becomes (-9)-5+(-28) =-9-5-28=-42 Hence the value is -42, so the correct option is (c).

The value of 66 − 2 × [10 + 2 × {18 − 2(7 − 4) × 3} ÷ 50] is:

A.46
B.42
C.54
D.41

Solution

BODMAS with brackets Open the brackets from the innermost one outwards:

()

, then

{}

, then

()

. (7-4)=3, so

2 \times 3 \times 3 = 18

{18 - 18} = 0

2 \times 0 \div 50 = 0

[10+0]=10

66 - 2 \times 10 = 66 - 20 = 46

Hence the value is 46, so the correct option is (a).

A man plans to travel a total distance of 60 km in 6 hours, covering part of the journey on foot and the rest by bicycle. His walking speed is

6 km / hr

, and his cycling speed is

12 km / hr

. How much distance (in km) does he cover on foot?

A.14
B.16
C.15
D.12

Solution

Speed, time and distance Let the time spent walking be t hours; then the cycling time is (6-t) hours. Total distance = walking distance + cycling distance: 6t+12(6-t)=60 6t+72-12t=60

- 6 t = - 12 \Rightarrow t = 2

hours Distance covered on foot

= 6 \times 2 = 12

Hence he walks 12 km, so the correct option is (d).

A vendor bought 120 shirts at the rate of ₹375 per shirt and sold them for ₹48,600. What is his gain/loss percentage ?

A.Gain 8%
B.Loss 12%
C.Loss 8%
D.Gain 12%

Solution

Profit percentage Cost price

= 120 \times 375 = ₹ 45000

Selling price =₹48600, and since SP > CP there is a gain. Gain =48600-45000=₹3600 Gain

% = \frac{3600}{45000} \times 100

= 8 %

Hence he gains

8 %

, so the correct option is (a).

If x varies directly as y and if x = 3, y = 5, then find the value of y when x = 7.8.

A.12
B.13
C.11
D.14

Solution

Direct variation In direct variation

x \propto y

, so the ratio

\frac{x}{y}

stays constant.

\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}}

\frac{3}{5} = \frac{7.8}{y}

y = \frac{7.8 \times 5}{3} = \frac{39}{3}

y=13 Hence the required value of y is 13, so the correct option is (b).

The price (per litre) of petrol increases by 40%. By what percent should its consumption be reduced such that the expenditure on it increases by 12% only ?

A.86%
B.20%
C.80%
D.18%

Solution

Price and consumption Expenditure = price

\times

consumption. Let the price be 100 and the consumption be 100, so expenditure

= 100 \times 100 = 10000

. New price

= 100 + 40 %

of 100=140 New expenditure

= 10000 + 12 %

of 10000=11200 New consumption

= \frac{11200}{140} = 80

Fall in consumption =100-80=20 Required reduction

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

Hence the consumption must be reduced by

20 %

, so the correct option is (b).

Mandar bought a calculator and sold it to Ketan at a profit of 5%. Ketan sold it to Tushar at a loss of 25%. If Tushar paid ₹3780, then the cost price (in ₹) of the calculator Mandar bought is

A.4800
B.4550
C.4600
D.4950

Solution

Successive profit and loss Let Mandar's cost price be ₹x. Mandar sells to Ketan at

5 %

profit: Ketan pays

x \times \frac{105}{100}

. Ketan sells to Tushar at

25 %

loss: Tushar pays

x \times \frac{105}{100} \times \frac{75}{100}

x \times \frac{105}{100} \times \frac{75}{100} = 3780

x \times \frac{63}{80} = 3780

x = 3780 \times \frac{80}{63} = 4800

Hence Mandar's cost price is ₹4800, so the correct option is (a).

If 4% of

x = 120

, then

x

is equal to:

A.3100
B.3000
C.6100
D.6000

Solution

Percentage of a number

4 %

of x=120

\frac{4}{100} \times x = 120

x = \frac{120 \times 100}{4}

x=3000 Hence x=3000, so the correct option is (b).

Amit invests a sum of ₹5400 and Gopal invests a sum of ₹10200 at the same rate of simple interest per annum. If, at the end of 4 years, Gopal gets ₹960 more interest than Amit, then find the rate of interest per annum (in percentage).

Solution

Simple interest rate Simple interest

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

, and here R and T are the same for both. So the extra interest comes only from the extra principal. Difference in principal =10200-5400=₹4800

\frac{4800 \times R \times 4}{100} = 960

192R=960

R = \frac{960}{192} = 5

Hence the rate of interest is

5 %

per annum, so the correct option is (b).

Find the effective discount for the scheme 'buy 10, get 6 free'.

A.60%
B.37.5%
C.73.5%
D.62.5%

Solution

Effective discount In 'buy 10, get 6 free' the customer pays for 10 articles but carries home 10+6=16 articles. So on 16 articles he is given 6 free - that is the discount. Effective discount

= \frac{6}{16} \times 100

= 37.5 %

The tempting mistake is

\frac{6}{10} \times 100 = 60 %

, but the discount must be compared with all 16 articles received. Hence the effective discount is

37.5 %

, so the correct option is (b).

The average weight of Akshay, Atul and Baban is 46 kg . If the average weight of Akshay and Atul be 39 kg and that of Atul and Baban be 46 kg , then the weight of Atul (in kg ) is:

A.42
B.52
C.32
D.47

Solution

Average of weights Total weight = average

\times

number of persons. Akshay + Atul + Baban

= 46 \times 3 = 138

kg Akshay + Atul

= 39 \times 2 = 78

kg Atul + Baban

= 46 \times 2 = 92

Baban =138-78=60 kg Atul =92-60=32 kg Hence the weight of Atul is 32 kg, so the correct option is (c).

Mukund has two grandsons Sachin and Chetan. 13 year old Sachin gets some money from Mukund's wealth and 14 year old Chetan gets rest of the money. But Sachin and Chetan will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate 10% compounded annually. When both turn 23 , they receive the same amount. How much had Mukund given Chetan (in ₹) initially, if total money with Mukund was ₹25200?

A.12000
B.11750
C.13200
D.13550

Solution

Compound interest division Sachin is 13, so his share grows for 23-13=10 years; Chetan is 14, so his grows for 23-14=9 years. Amount

= P {(1 + \frac{10}{100})}^{n} = P {(\frac{11}{10})}^{n}

Let the shares be S and C. Equal amounts at age 23 give:

S {(\frac{11}{10})}^{10} = C {(\frac{11}{10})}^{9}

\frac{S}{C} = {(\frac{11}{10})}^{9 - 10} = \frac{10}{11}

, so S:C=10:11.

10+11=21 parts

= ₹ 25200 \Rightarrow 1

part

= \frac{25200}{21} = 1200

Chetan's share

= 11 \times 1200 = 13200

Hence Mukund gave Chetan ₹13200, so the correct option is (c).

Evaluate:

32 \div 8 \times 3 - 2 \times 3

Solution

BODMAS By BODMAS,

\div

and

\times

are done before -, and between themselves from left to right.

32 \div 8 = 4

4 \times 3 = 12

2 \times 3 = 6

12-6=6 Hence the value is 6, so the correct option is (a).

Find the area (in

{cm}^{2}

) of the card board that is required to make a box of size

12 cm \times 14 cm \times 8 cm

A.736
B.752
C.748
D.764

Solution

Surface area of a cuboid The card board covers all six faces, so we need the total surface area of the cuboid.

Total surface area =2(lb+bh+hl) Here l=14, b=12 and h=8:

= 2 (14 \times 12 + 12 \times 8 + 8 \times 14)

=2(168+96+112)

= 2 \times 376 = 752

Hence

752 {c m}^{2}

of card board is required, so the correct option is (b).

The greatest number, which divides 711 and 1040 to leave 48 and 20 respectively as remainders, is:

A.71
B.32
C.63
D.51

Solution

HCF with remainders If a number leaves remainder r on dividing a, then it divides (a-r) exactly. 711-48=663 1040-20=1020 The greatest such number is the HCF of 663 and 1020.

663 = 3 \times 13 \times 17

1020 = 2^{2} \times 3 \times 5 \times 17

HCF

= 3 \times 17 = 51

51 is larger than both remainders 48 and 20, so the division is valid. Hence the required number is 51, so the correct option is (d).

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