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SSC GD Constable 10 February 2025 · Shift 2 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 68 of 210

10 February 2025 · Shift 2

Elementary Mathematics

20 questions ~20 min readFree

20 questions

Find the value of

((19 \times 3) \times (8 \div 2 \times \frac{(18 - 14)}{4}))

A.245
B.214
C.228
D.226

Solution

Order of operations (BODMAS) Open the innermost bracket first, and inside every bracket follow BODMAS: division and multiplication before addition and subtraction. Inside the curly bracket, the fraction bar acts as a bracket:

\frac{18 - 14}{4} = \frac{4}{4} = 1

So the curly bracket

= 8 \div 2 \times 1 = 4 \times 1 = 4

The square bracket now reads

(19 \times 3) \times 4 = 57 \times 4

57 \times 4 = 228

So the correct option is (c).

Find the value of

((12 \div 4) \times (\frac{15}{5} + \frac{16}{1} \times (9 - 8)))

A.57
B.44
C.69
D.61

Solution

Simplification of brackets Simplify the curly bracket first; inside it multiplication and division are done before the addition.

\frac{15}{5} = 3

\frac{16}{1} \times (9 - 8) = 16 \times 1 = 16

Curly bracket =3+16=19

(12 \div 4) = 3

, so the whole expression

= 3 \times 19

3 \times 19 = 57

So the correct option is (a).

Find the mean proportional of

(12 + 6 \sqrt{2})

and

(8 - 4 \sqrt{2})

A. $4 \sqrt{3}$
B. $3 \sqrt{2}$
C. $3 \sqrt{3}$
D. $4 \sqrt{2}$

Solution

Mean proportional of surds The mean proportional m of a and b satisfies a:m=m:b, so

m^{2} = a b

and

m = \sqrt{a b}

a b = (12 + 6 \sqrt{2}) (8 - 4 \sqrt{2})

= 96 - 48 \sqrt{2} + 48 \sqrt{2} - 24 \times 2

The two surd terms cancel, leaving ab=96-48=48.

m = \sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3}

So the correct option is (a).

The population of a district is 360000, out of which 225000 are males. 35% of the population is literate. If 17% males are literate, then what percentage of females are literate ?

A.68%
B.67%
C.63%
D.65%

Solution

Percentage of literacy Count the literates as numbers first; percentages of two different groups can never be subtracted directly. Number of females =360000-225000=135000 Total literates

= \frac{35}{100} \times 360000 = 126000

Literate males

= \frac{17}{100} \times 225000 = 38250

Literate females =126000-38250=87750

Required percentage

= \frac{87750}{135000} \times 100 = \frac{87750}{1350}

= 65 %

So the correct option is (d).

The average age of 57 students of a class is 13 years. If the age of the teacher is also included, the average age of the whole group becomes 14 years. The age (in years) of the teacher is:

A.70
B.69
C.72
D.71

Solution

Average age of a group Total = average

\times

number of members, so convert both averages into totals. Total age of 57 students

= 57 \times 13 = 741

years With the teacher the group has 58 members: total

= 58 \times 14 = 812

years

Age of the teacher =812-741=71 years Quick check: the teacher raises the average of 58 people by 1 year, so his age

= 14 + 57 \times 1 = 71

. So the correct option is (d).

Applied to a bill of ₹600, the difference between a single discount of

25 %

and two successive discounts of

10 %

and

15 %

is:

A.₹1.5
B.₹11
C.₹10
D.₹9

Solution

Single vs successive discount Successive discounts are never added: the second one acts on an amount that has already been reduced. After a single discount of

25 %

600 \times \frac{75}{100} = 450

After

10 %

600 \times \frac{90}{100} = 540

Then after

15 %

540 \times \frac{85}{100} = 459

Difference =459-450=9 So the difference is ₹9 and the correct option is (d).

A car covers a distance of 30 km and 250 metres in 25 minutes. The speed of the car in km per hour is

____

A.76.2
B.67.2
C.72.6
D.62.7

Solution

Speed from distance and time Bring the distance to kilometres and the time to hours before dividing. Distance =30 km 250 m

= 30 + \frac{250}{1000} = 30.25

km Time =25 minutes

= \frac{25}{60}

hour Speed

= \frac{distance}{time} = 30.25 \times \frac{60}{25}

= 30.25 \times 2.4 = 72.6

So the speed is 72.6 km/h and the correct option is (c).

33 guava trees, 55 banana trees and 88 malta trees need to be planted in rows such that each row contains the same number of trees of one type only. The minimum number of rows in which the trees may be planted is:

A.18
B.12
C.20
D.16

Solution

HCF in row arrangement Every row holds the same number of trees, so that number must divide 33, 55 and 88; to get the fewest rows it must be the largest such divisor, the HCF.

33 = 3 \times 11

55 = 5 \times 11

88 = 8 \times 11

, so HCF =11. Guava rows

= \frac{33}{11} = 3

Banana rows

= \frac{55}{11} = 5

Malta rows

= \frac{88}{11} = 8

Total rows =3+5+8=16 So the correct option is (d).

Sachin has two grandsons Chetan and Vipul. 13 year old Chetan gets some money from Sachin's wealth and 14 year old Vipul gets rest of the money. But Chetan and Vipul will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate

8 %

compounded annually. When both turn 23 , they receive the same amount. How much had Sachin given Vipul (in ₹) initially, if total money with Sachin was ₹23400?

A.11000
B.12150
C.11250
D.12500

Solution

Compound interest sharing Chetan's money grows for 23-13=10 years and Vipul's for 23-14=9 years, and the two maturity amounts are equal. Amount

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

, and here

1 + \frac{8}{100} = \frac{27}{25}

C {(\frac{27}{25})}^{10} = V {(\frac{27}{25})}^{9}

Cancelling

{(\frac{27}{25})}^{9}

from both sides:

C \times \frac{27}{25} = V

\frac{C}{V} = \frac{25}{27}

- the boy who gets one extra year of growth needs one factor less to start with.

Total parts =25+27=52, and these 52 parts stand for ₹23,400. 1 part

= \frac{23400}{52} = 450

Vipul's share

= 27 \times 450 = 12150

So Vipul was given ₹12,150 and the correct option is (b).

The speed of a stream is

4 km / h

. A boat can go 20 km downstream and 14 km upstream in 3 hours. What is the speed (in

km / h

) of the boat in still water?

A.12
B.9
C.6
D.5

Solution

Boat and stream Let the speed of the boat in still water be x km/h; the stream is 4 km/h. Downstream speed =x+4 and upstream speed =x-4.

\frac{20}{x + 4} + \frac{14}{x - 4} = 3

20 (x - 4) + 14 (x + 4) = 3 (x^{2} - 16)

34 x - 24 = 3 x^{2} - 48

, so

3 x^{2} - 34 x - 24 = 0

3 x^{2} - 36 x + 2 x - 24 = 0 \Rightarrow (x - 12) (3 x + 2) = 0

A speed cannot be negative, so x=12. So the correct option is (a).

If at the same rate of interest, in 2 years, the simple interest is ₹44 and compound interest is ₹46, then what is the principal (in ₹) ?

A.237
B.235
C.246
D.242

Solution

SI and CI difference For 2 years at the same rate, the whole gap between CI and SI is one year's interest on the first year's interest. SI for 2 years =44, so the interest of one year

= \frac{44}{2} = 22

. CI-SI=46-44=2

This 2 is one year's interest on 22:

\frac{22 \times R}{100} = 2 \Rightarrow R = \frac{100}{11} %

S I = \frac{P \times R \times T}{100} \Rightarrow 44 = \frac{P \times \frac{100}{11} \times 2}{100}

44 = \frac{2 P}{11} \Rightarrow P = \frac{44 \times 11}{2} = 242

So the principal is ₹242 and the correct option is (d).

Prakhar goes from his home to school at a speed of

24 km / hr

and returns at the speed of

36 km / hr

. Find the average speed for the whole journey.

A. $31.6 km / hr$
B. $28.8 km / hr$
C. $27.9 km / hr$
D. $26.3 km / hr$

Solution

Average speed both ways The two legs cover the same distance, so the average speed is the harmonic mean of the speeds, never their simple average. Average speed

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

, with x=24 and y=36.

= \frac{2 \times 24 \times 36}{24 + 36} = \frac{1728}{60}

=28.8 So the average speed is 28.8 km/hr and the correct option is (b).

A dishonest shopkeeper promises to sell his goods at cost price. However, he uses a weight that actually weighs 39% less than what is written on it. Find his profit percentage.

A. $62 \frac{57}{61}$ %
B. $65 \frac{114}{61} %$
C. $63 \frac{57}{61} %$
D. $64 \frac{58}{61} %$

Solution

Profit by false weight He sells at cost price, so his whole profit comes from the short weight: he is paid for 1000 g but hands over less. Weight actually given

= 1000 - 39 %

of 1000=1000-390=610 g Take the cost price as ₹1 per gram, so goods worth ₹610 are sold for ₹1000.

Profit =1000-610=390 Profit

% = \frac{profit}{cost price} \times 100 = \frac{390}{610} \times 100

= \frac{3900}{61} = 63 \frac{57}{61} %

So the correct option is (c).

2 %

x = 240

, then

x

is equal to:

A.12000
B.24100
C.12100
D.24000

Solution

Basic percentage equation "

2 %

of x" means

\frac{2}{100} \times x

\frac{2 x}{100} = 240

x = \frac{240 \times 100}{2}

x=12000 So the correct option is (a).

Compare the ratios 2 : 3, 5 : 7, 3 : 4, 5 : 9. Which one is the smallest?

A.5 : 9
B. $3 : 4$
C.2 : 3
D.5 : 7

Solution

Comparison of ratios Ratios can only be compared after each is written as a single value - turn every ratio into a decimal.

\frac{2}{3} = 0.667

\frac{5}{7} = 0.714

\frac{3}{4} = 0.750

\frac{5}{9} = 0.556

The smallest value is 0.556, which belongs to 5:9. So the correct option is (a).

Find the volume of a hemisphere of radius

6 \sqrt{3} cm

A. $452 \sqrt{3} π {cm}^{3}$
B. $442 \sqrt{3} π {cm}^{3}$
C. $432 \sqrt{3} π {cm}^{3}$
D. $462 \sqrt{3} π {cm}^{3}$

Solution

Volume of a hemisphere Volume of a hemisphere

= \frac{2}{3} π r^{3}

, which is half the sphere's

\frac{4}{3} π r^{3}

Here

r = 6 \sqrt{3}

, so

r^{3} = 6^{3} \times {(\sqrt{3})}^{3} = 216 \times 3 \sqrt{3} = 648 \sqrt{3}

Volume

= \frac{2}{3} \times π \times 648 \sqrt{3}

= 2 \times 216 \sqrt{3} π = 432 \sqrt{3} π {cm}^{3}

So the correct option is (c).

The cost price of a calculator is ₹315. If the profit percentage is 20%, then what is the value (in ₹) of the profit ?

A.71
B.63
C.60
D.68

Solution

Profit from profit percentage Profit percentage is always reckoned on the cost price, so take

20 %

of ₹315. Profit

= \frac{20}{100} \times 315

= \frac{315}{5} = 63

So the profit is ₹63 and the correct option is (b).

Keshav invests a sum of ₹5400 and Tushar invests a sum of ₹10200 at the same rate of simple interest per annum. If, at the end of 3 years, Tushar gets ₹360 more interest than Keshav, then find the rate of interest per annum (in percentage).

A.1.5
B.3.5
C.2.5
D.4.5

Solution

Simple interest rate Both sums run at the same rate for the same 3 years, so the extra interest is produced only by the extra principal. Extra principal =10200-5400=4800

S I = \frac{P \times R \times T}{100} \Rightarrow 360 = \frac{4800 \times R \times 3}{100}

360=144R

R = \frac{360}{144} = 2.5

So the rate is

2.5 %

per annum and the correct option is (c).

A, B and C can complete a piece of work in 12, 20 and 15 days, respectively. A, B and C will complete the work while working on alternate days. C works on the

1^{st}

day, B works on the

2^{nd}

day and A works on the

3^{rd}

day and so on. In how many days will the work get complete ?

A.24
B.15
C.30
D.12

Solution

Work on alternate days Take the total work as the LCM of 12, 20 and 15, i.e. 60 units, so that each day's work is a whole number. A does

\frac{60}{12} = 5

, B does

\frac{60}{20} = 3

and C does

\frac{60}{15} = 4

units per day. The order given is C, then B, then A, so one cycle of 3 days does 4+3+5=12 units.

Number of full cycles needed

= \frac{60}{12} = 5

Days taken

= 5 \times 3 = 15

The work ends exactly at the close of the 5th cycle, so it is finished in 15 days. So the correct option is (b).

LCM and HCF of two numbers are 154 and 7 respectively. If one of the numbers is 77 , then find the other number.

A.16
B.15
C.13
D.14

Solution

LCM and HCF relation For any two numbers, LCM

\times

HCF = product of the two numbers.

154 \times 7 = 77 \times x

1078=77x

x = \frac{1078}{77} = 14

So the other number is 14 and the correct option is (d).

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