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SSC GD Constable 21 February 2025 · Shift 2 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 188 of 210

21 February 2025 · Shift 2

Elementary Mathematics

20 questions ~20 min readFree

20 questions

If the average marks of three sections of Class X, having 35, 40, and 25 students, respectively, are 90,88 , and 80 , then the average marks of all the students of Class X, is:

A.76.8
B.78.6
C.87.6
D.86.7

Solution

Weighted average Total marks of a section = number of students × average marks of that section.

Section totals:

35 \times 90 = 3150

40 \times 88 = 3520

25 \times 80 = 2000

Total marks =3150+3520+2000=8670 and total students =35+40+25=100 Required average

= \frac{8670}{100} = 86.7

So option (d) is correct.

Let

x

be the greatest number which, when divided by 31, 61 and 91, leaves the same remainder. The LCM of x and 48 is:

A.360
B.178
C.240
D.420

Solution

HCF and LCM The greatest number that leaves the same remainder with 31, 61 and 91 is the HCF of their differences - the common remainder cancels in every difference.

x = H C F (61 - 31, 91 - 61, 91 - 31) = H C F (30, 30, 60) = 30

Now factorise:

30 = 2 \times 3 \times 5

and

48 = 2^{4} \times 3

L C M (30, 48) = 2^{4} \times 3 \times 5 = 240

So option (c) is correct.

When

x

is subtracted from each of 55, 50, 23 and 22, the resulting numbers, in this order, are in proportion. What is the fourth proportion of 3, 7 and x ?

A.35
B.32
C.38
D.36

Solution

Proportion In proportion the product of the extremes equals the product of the means:

\frac{55 - x}{50 - x} = \frac{23 - x}{22 - x}

(55-x)(22-x)=(50-x)(23-x)

1210 - 77 x + x^{2} = 1150 - 73 x + x^{2}

1210 - 1150 = 77 x - 73 x \Rightarrow 60 = 4 x \Rightarrow x = 15

Fourth proportional of 3, 7 and 15

= \frac{7 \times 15}{3} = 35

So option (a) is correct.

A person drives 60 km at a speed of 30 km/hr, then 30 km at 60 km/hr, and finally 45 km at 90 km/hr. What is the average speed (in

km / hr

) for the entire trip ?

A.50
B.42.5
C.45
D.47.5

Solution

Average speed Average speed

= \frac{total distance}{total time}

- never the average of the three speeds.

Times taken:

\frac{60}{30} = 2

\frac{30}{60} = 0.5

\frac{45}{90} = 0.5

h Total distance =60+30+45=135 km, total time =2+0.5+0.5=3 h Average speed

= \frac{135}{3} = 45

km/hr So option (c) is correct.

The average weight of Rakesh, Shivam and Rupesh is 46 kg . If the average weight of Rakesh and Shivam be 40 kg and that of Shivam and Rupesh be 48 kg , then the weight of Shivam (in kg) is:

A.38
B.53
C.48
D.58

Solution

Average of weights Sum of weights = average × number of persons.

R + S + P = 46 \times 3 = 138

R + S = 40 \times 2 = 80

and

S + P = 48 \times 2 = 96

Adding the last two: (R+S)+(S+P)=80+96=176, which counts Shivam twice. S=176-138=38 So option (a) is correct.

The population of a district is 383000, out of which 221000 are males. 58% of the population is literate. If 58% males are literate, then what percentage of females are literate ?

A.57%
B. $61 %$
C.58%
D.56%

Solution

Percentage literacy Females =383000-221000=162000 Literate people

= \frac{58}{100} \times 383000 = 222140

Literate males

= \frac{58}{100} \times 221000 = 128180

Literate females =222140-128180=93960 Required percentage

= \frac{93960}{162000} \times 100 = 58 %

Shortcut: the whole population and the male part have the SAME literacy rate, so the remaining (female) part must also be at that rate - 58%. So option (c) is correct.

400 custard apples were bought at ₹1200 per hundred and were sold at a profit of ₹900. Find the selling price (in ₹) per dozen of custard apples.

A.171
B.181
C.161
D.186

Solution

Profit and selling price 400 apples are 4 hundreds, so cost price

= 4 \times 1200 = 4800

, i.e. ₹4,800. Selling price = cost price + profit =4800+900=5700, i.e. ₹5,700. Selling price of one apple

= \frac{5700}{400} = 14.25

Selling price per dozen

= 14.25 \times 12 = 171

So the price per dozen is ₹171 and option (a) is correct.

Find the value of

((26 \times 14) \times (9 \div 3 \times \frac{(19 - 15)}{2}))

A.2184
B.2165
C.2197
D.2188

Solution

BODMAS simplification By BODMAS the innermost bracket comes first: 19-15=4, so

\frac{19 - 15}{2} = 2

Inside the braces, divide before multiplying:

9 \div 3 \times 2 = 3 \times 2 = 6

The first bracket:

26 \times 14 = 364

Now multiply:

364 \times 6 = 2184

So option (a) is correct.

The value of 22 − 2 × [7 + 8 × {14 − 7(6 − 5) × 2} ÷ 37] is:

A.18
B.8
C.7
D.2

Solution

BODMAS simplification Start with the innermost bracket: 6-5=1 Now the braces:

14 - 7 \times 1 \times 2 = 14 - 14 = 0

The braces vanish, so

8 \times 0 \div 37 = 0

and the square bracket becomes 7+0=7

22 - 2 \times 7 = 22 - 14 = 8

So option (b) is correct.

Evaluate:

16 + 8 \div 4 - 3 \times 3

A.8
B.11
C.12
D.9

Solution

BODMAS simplification By BODMAS, division and multiplication are done before addition and subtraction.

8 \div 4 = 2

and

3 \times 3 = 9

16+2-9=18-9=9 So option (d) is correct.

A man sold an article for ₹221, there by gaining 10.5%. The cost of the article (in ₹) was:

A.247
B.244
C.200
D.198

Solution

Cost price from gain% A gain of 10.5% means the selling price is 110.5% of the cost price.

221 = C P \times \frac{110.5}{100}

C P = \frac{221 \times 100}{110.5} = \frac{22100}{110.5} = 200

So the cost price is ₹200 and option (c) is correct.

If the height of a cylinder is 16 cm and radius is 14 cm , what is the curved surface area (in

{cm}^{2}

) of it ?

A.1424
B.1418
C.1408
D.1412

Solution

Curved surface of a cylinder The curved surface is only the side of the cylinder - rolled flat it is a rectangle of length

2 π r

and width h.

Curved surface area

= 2 π r h

= 2 \times \frac{22}{7} \times 14 \times 16

= 2 \times 22 \times 2 \times 16 = 1408

So the curved surface area is

1408 c m^{2}

and option (c) is correct.

The area of a triangle is constant. Find the percentage increase in its perpendicular height if the base of the triangle is decreased by 20%.

A.12.5%
B.15%
C.20%
D.25%

Solution

Percentage change, fixed area Area

= \frac{1}{2} \times b a s e \times h e i g h t

, so with the area fixed, base and height are inversely proportional. Take base =5 and height =4 units, so area =10 square units. New base

= 5 \times \frac{80}{100} = 4

units.

New height

= \frac{2 \times 10}{4} = 5

units. Increase

= \frac{5 - 4}{4} \times 100 = 25 %

So option (d) is correct.

When difference between compound and simple interest for three years is ₹152 at 4% interest per annum, the principal is ₹

____

A.32450
B.30675
C.31250
D.31670

Solution

CI − SI for three years For three years,

C I - S I = P \times \frac{R^{2} (300 + R)}{100^{3}}

152 = P \times \frac{4^{2} (300 + 4)}{1000000}

152 = P \times \frac{16 \times 304}{1000000} = \frac{4864 P}{1000000}

P = \frac{152 \times 1000000}{4864} = \frac{1000000}{32} = 31250

So the principal is ₹31,250 and option (c) is correct.

Find the fourth proportional of 4, 6 and 5.

A.9.5
B.8
C.7.5
D.6

Solution

Fourth proportional If a:b::c:d, the fourth proportional is

d = \frac{b \times c}{a}

Here a=4, b=6, c=5, so

d = \frac{6 \times 5}{4} = \frac{30}{4} = 7.5

So option (c) is correct.

Two trains having lengths of 110 m and 390 m are running at speeds of 120

km / h

and

130 km / h

, respectively, in the same direction. The time taken (in minutes) by the faster train, coming from behind, to completely cross the other train is:

A.3
B.2
C.1.8
D.5

Solution

Trains crossing, same direction Both trains move the same way, so the speeds subtract: relative speed =130-120=10 km/h.

10 k m / h = 10 \times \frac{5}{18} = \frac{25}{9}

m/s

To cross completely, the faster train must cover both lengths: 390+110=500 m. Time

= \frac{500}{\frac{25}{9}} = \frac{500 \times 9}{25} = 180

seconds

= \frac{180}{60} = 3

minutes So option (a) is correct.

Parth, Rizwan and Sangeeta can do a piece of work in 40,60 and 80 days, respectively. In how many days can Parth do the work if he is assisted by Rizwan and Sangeeta on every third day?

A. $28 \frac{4}{5}$
B. $29 \frac{4}{5}$
C. $28 \frac{3}{13}$
D. $29 \frac{3}{13}$

Solution

Time and work, help every third day Let the whole work be

L C M (40, 60, 80) = 240

units. One day's work: Parth

= \frac{240}{40} = 6

, Rizwan

= \frac{240}{60} = 4

, Sangeeta

= \frac{240}{80} = 3

units.

A 3-day cycle finishes 6+6+(6+4+3)=25 units.

25 \times 9 = 225

units are done in

3 \times 9 = 27

days. Remaining =240-225=15 units; on days 28 and 29 Parth alone does 6+6=12 units. Left =15-12=3 units, and day 30 is a 'third day' worth 13 units, needing

\frac{3}{13}

of a day. Total time

= 27 + 2 + \frac{3}{13} = 29 \frac{3}{13}

days So option (d) is correct.

The marked price of a wrist watch is ₹1,200. It is sold at two successive discounts of 5% and 8%, respectively. Find its selling price.

A.₹1,048.80
B.₹1,052.60
C.₹1,070.85
D.₹1,085.65

Solution

Successive discounts Successive discounts multiply, they do not add - the second 8% is taken on the reduced price.

S P = 1200 \times \frac{95}{100} \times \frac{92}{100}

= 1140 \times \frac{92}{100} = 1048.80

So the selling price is ₹1,048.80 and option (a) is correct.

Shan has ₹1617 with him. He divided it amongst his sons Piyush and Manoj and asked them to invest it at 10% rate of interest compounded annually. It was seen that Piyush and Manoj got same amount after 11 and 12 years respectively. How much (in ₹) did Shan give to Piyush ?

A.847
B.697
C.870
D.770

Solution

Compound interest, equal amounts Let Piyush get P and Manoj get M; their amounts are equal after 11 and 12 years.

P {(1 + \frac{10}{100})}^{11} = M {(1 + \frac{10}{100})}^{12}

\frac{P}{M} = {(\frac{11}{10})}^{12 - 11} = \frac{11}{10}

So P:M=11:10 and 11+10=21 units correspond to ₹1617. 1 unit

= \frac{1617}{21} = 77

Piyush's share

= 11 \times 77 = 847

So Shan gave Piyush ₹847 and option (a) is correct.

The simple interest on a principal amount (in ₹) is ₹168 for a period of 8 years at the rate of

4 %

per annum. The principal amount (in ₹) is:

A.517
B.527
C.534
D.525

Solution

Simple interest

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

, so

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

P = \frac{168 \times 100}{4 \times 8}

P = \frac{16800}{32} = 525

So the principal is ₹525 and option (d) is correct.

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