Quantitative Aptitude Questions for SSC CHSL Set-2 download PDF based on previous year question paper of SSC exams. 15 Very important Quantitative Aptitude questions for SSC CHSL Exam.
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Question 1: In what ratio salt costing Rs.13/kg mixed with another salt costing Rs.7/ kg to get a mixture costing Rs.9/kg?
a) 4 : 7
b) 3 : 5
c) 2 : 3
d) 1 : 2
Question 2: $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$ is approximately equal to ___________ .
a) 1.5
b) 2
c) 2.5
d) 3
Question 3: The sum of the present ages of a father and his son is 126 years. 8 years ago their respective ages were in the ratio of 7 : 4. After 7 years what will be the ratio of ages of father and son?
a) 23 : 19
b) 17 : 11
c) 27 : 20
d) 34 : 23
Question 4: A bus can travel at 150 km/hr without stoppages and at 120 km/hr with stoppages. How many minutes per hour does the bus stop?
a) 15 minutes
b) 12 minutes
c) 16 minutes
d) 20 minutes
Question 5: A train of length 120 m crossed a platform in 36 seconds. It crosses a man standing on the platform in 24 seconds running at the same speed. Find the length of the platform.
a) 120 m
b) 60 m
c) 104 m
d) 72 m
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Question 6: A train crosses a pole in 12 seconds and a platform of length 160 m in 32 seconds running at the same speed. Find the length of the train.
a) 108 m
b) 96 m
c) 120 m
d) 84 m
Question 7: ΔXYZ is right angled at Y. If $sinX = \frac{4}{5}$, then what is the value of $cos Z$ ?
a) $\frac{3}{4}$
b) $\frac{5}{3}$
c) $\frac{4}{5}$
d) $\frac{4}{3}$
Question 8: What is the value of $(\frac{1}{3} – sec 45^\circ)$ ?
a) $\frac{(4-\sqrt3)}{2}$
b) $\frac{(\sqrt{6}-1)}{\sqrt{3}}$
c) $\frac{(1-3\sqrt2)}{3}$
d) $\frac{(\sqrt2-1)}{\sqrt{2}}$
Question 9: If θ is positive acute angle Find whether the equation 3 $(\sec^{2}\theta-\tan^{2}\theta)=5$ is true or not ?
a) True
b) False
c) can not be determined
d) None of these
Question 10: What is the absolute difference of the roots of the equation $2x^{2}-6x+8 = 0$ ?
a) $\sqrt{6}i$
b) $\sqrt{5}i$
c) $\sqrt{7}i$
d) $\sqrt{3}i$
Question 11: A class has two sections A and B. The average marks of two Sections in an examination is 75. The average marks of Section A is 72 and that of Section B is 80. Find the ratio of the number of students in Section A to Section B
a) $5 : 3$
b) $3 : 5$
c) $9 : 10$
d) $10 : 9$
Question 12: If ‘a’ is a positive integer such that $a^2=2a+3$ then what is the value of $a^3$?
a) 1
b) 8
c) 27
d) 64
Question 13: If $\log{2} = 0.301$ and $log{3}=0.477$, find the value of $log{72}$.
a) 1.424
b) 1.857
c) 1.987
d) 2.357
Question 14: If $\log{16}=1.2$ then what is the value of $\log{64}$?
a) 1.6
b) 1.7
c) 1.8
d) 1.9
Question 15: If $log_{x}y = 5 $ and $log_{x}729= 6$ then find the value of y.
a) 625
b) 243
c) 81
d) 343
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Answers & Solutions:
1) Answer (D)
Let quantity of Rs. 13/kg and Rs. 7/kg salt mixed be $x$ and $y$ kg respectively.
According to ques,
=> $13x+7y=9(x+y)$
=> $13x+7y=9x+9y$
=> $13x-9x=9y-7y$
=> $4x=2y$
=> $\frac{x}{y}=\frac{2}{4}=\frac{1}{2}$
$\therefore$ Required ratio = $1:2$
=> Ans – (D)
2) Answer (B)
Expression = $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
= $(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6})-(\frac{1}{8})$
The first term above is a geometric progression with first term, $a=1$ and common ratio, $r=\frac{1}{2}$
Number of terms, $n=7$
Sum of $n$ terms of G.P. (when $r<1$) = $\frac{a(1-r^n)}{1-r}$
= $[\frac{1(1-\frac{1}{2}^7)}{1-\frac{1}{2}}]-(\frac{1}{8})$
= $[2\times(1-\frac{1}{128})]-(\frac{1}{8})$
= $\frac{127}{64}-\frac{8}{64}$
= $\frac{119}{64}=1.85\approx2$
=> Ans – (B)
3) Answer (B)
Sum of the present ages of a father and his son = 126 years
Sum of their ages 8 years ago = $126-8(2)=110$ years
=> Father’s age at that time = $\frac{7}{(7+4)}\times110=70$ years
=> Son’s age at that time = $110-70=40$ years
Thus, present age of father = $70+8=78$ years
=> Son’s present age = $40+8=48$ years
$\therefore$ Ratio of ages of father and son after 7 years = $\frac{78+7}{48+7}=\frac{85}{55}$
= $17:11$
=> Ans – (B)
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4) Answer (B)
Due to stoppages, bus travelled 30 km less in an hour.
Time taken to travel 30 km $= \dfrac{30}{150}\times60 = 12$ minutes
5) Answer (B)
Speed of the train to cross the man = 120/24 = 5 m/sec
Speed of the train to cross the platform = (120+P)/36
$\dfrac{120+P}{36} = 5$
⇒ $120+P = 180$
⇒ $P = 60$
6) Answer (B)
Let the length of the train be ‘T’ metres.
Speed of the train to cross the pole in 12 seconds = T/12 m/sec
Speed of the train to cross the platform in 32 seconds = (T+160)/32 m/sec
Here, Speeds are equal.
⇒ $\dfrac{T}{12} = \dfrac{T+160}{32}$
⇒ $8T = 3T + 480$
⇒ $5T = 480$
⇒ $T = 96 m$
Hence, The Length of the train = 96 m
7) Answer (C)
Given : $\sin X$ = $\frac{4}{5}$
Also, $\sin X=\frac{YZ}{XZ}=\frac{4}{5}$
Let YZ = 4 cm and XZ = 5 cm
To find : $\cos Z=\frac{YZ}{XZ}$
= $\frac{4}{5}$
=> Ans – (C)
8) Answer (C)
Expression : $(\frac{1}{3} – sec 45^\circ)$
= $\frac{1}{3}-\sqrt2$
= $\frac{(1-3\sqrt2)}{3}$
=> Ans – (C)
9) Answer (B)
Given that : 3 $(\sec^{2}\theta-\tan^{2}\theta)=5$
we know that 1 + $tan^2 \theta = sec^2 \theta$
using the above identity ,
L.H.S :: 3 $(\sec^{2}\theta-\tan^{2}\theta)$ = 3 $(1 + tan^2 \theta – tan^2 \theta)$ = 3
R.H.S :: 5
As , L.H.S =/= R.H.S the given equation is false
10) Answer (C)
Let a and b be roots of the equation.
a+b = $\frac{6}{2}$ = $3$
ab = $\frac{8}{2}$ = $4$
$(a-b)^{2}$ = $(a+b)^{2}$ -$4ab$ = 9 -4(4) = -7
$a-b$ = $\sqrt{7} * \sqrt {-1}$ = $\sqrt{7}i$
11) Answer (A)
Let the number of students in Section A and Section B be $x$ and $y$ respectively
Total marks of the students from Section A = $72x$
Total marks of the students from Section B = $80y$
$\therefore 72x + 80y = 75(x+y)$
$\implies 5y = 3x$
$ \implies \frac{x}{y}=\frac{5}{3}$
Hence , the correct option is A
12) Answer (C)
The given equation can be written as,
$a^2-2a-3=0$
=>$(a-3)(a+1)=0$
As it is given that ‘a’ is a positive integer we get, a = 3.
So we get, $a^3=3^3=27$
13) Answer (B)
$72 = 2^3 \times 3^2$
=> $log{72} = log{2^3 \times 3^2} = log{2^3} + log{3^2} = 3log{2} + 2log{3} = 3\times 0.301 + 2 \times 0.477 = 1.857$
14) Answer (C)
It is given that,
$\log{16}=1.2$
$=>log{2^4}=1.2$
$=>4log{2}=1.2$
$=>log{2}=0.3$
Now let us find out the value of $\log{64}$
$\log{64}=log{2^6}=6log{2}=6*0.3=1.8$
15) Answer (B)
$log_{x}y = 5 $ and $log_{x}729= 6$
=> $x^5 = y$ and $x^6 = 729$ => $x = 729^{1/6} = (3^6)^{1/6} = 3$
Hence
$ y = 3^5 = 243$
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