Latest SSC jobs   »   Important Quantitative Aptitude Questions For SSC CHSL And CPO 2018 : 22nd March 2018 (with Solutions)

# Important Quantitative Aptitude Questions For SSC CHSL And CPO 2018 : 22nd March 2018 (with Solutions)

Dear students, you know that QUANT is a part of getting points and every chapter is important. Therefore, we are providing 15 questions of quant. Solve all these quizzes every day so that you can improve your accuracy and speed. We also provide lots of quant questions. So you can practice that chapter which takes more time to solve the questions.
प्रिय पाठकों, आप सभी जानते हैं कि संख्याताम्क अभियोग्यता का भाग बहुत ही महत्वपूर्ण है. इसलिए हम आपको संख्यात्मक अभियोग्यता कि 15 प्रश्नों कि प्रश्नोत्तरी प्रदान कर रहे हैं. इन सभी प्रश्नोत्तरी को दैनिक रूप से हल कीजिये ताकि आप अपनी गति और सटीकता में वृद्धि कर सकें. हम आपको अन्य कई संख्यात्मक अभियोग्यता के प्रश्न प्रदान करेंगे. ताकि आप पाठ्यक्रम अनुसार उन्हें हल कर पायें.

Q1. There is a small island in the middle of a 100 metre wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks and in line with the tree. If the angles of elevation of the top of tree from P and Q are respectively 30° and 45°, find the height of the tree.
(a) 35.6 metre
(b) 36.6 metre
(c) 37.6 metre
(d) 46.6 metre
Ans.(b)
Sol.

In ∆APO
tan⁡30°=h/PO
1/√3=h/PO
PO = h√3 …(i)

In ∆AOQ
tan⁡45°=h/OQ
1=h/OQ
OQ = h  …(ii)

PO + OQ = h√3 + h
PQ = h(√3 + 1)
100/(√3+1)=h
h=100/(1.73+1)=100/2.73
= 36.6

Q2. Two circles with centres A and B of radii 3 cm and 4 cm respectively intersect at two points C and D such that AC and BC are tangents to the two circles. Find the length of common chord CD.
(a) 4.4 cm
(b) 4.3 cm
(c) 4.5 cm
(d) 4.8 cm
Ans.(d)
Sol.

In ∆ACB
AB² = AC² + BC²
AB² = 4² + 3²
AB = 5
Let AP = x, BP = 5 – x & CP = PD = y

In ∆ACP
3² = x² + y² …(i)
In ∆CPB
4² = (5 – x)² + y²
16 = 25 + x² – 10x + y² …(ii)

Subtracting (ii)  – (i)
16 – 9 = 25 + x² – 10x + y² – x² – y²
7 = 25 – 10x
10x = 18
x = 1.8
9 = (1.8)² + y²
y² = 9 – 3.24
y² = 5.76, y = 2.4
CD = 2y = 2 × 2.4 = 4.8

Q3. The value of 3(sin x – cos x)^4+6(sin⁡x+cos⁡x )^2+4(sin^6⁡x+cos^6⁡x ) is
(a) 13
(b) 24
(c) 0
(d) 9
Ans.(a)
Sol. 3[(sin x – cos x)²]²
= 3[sin² x + cos² x – 2 sin x cos x]²
= 3(1 – 2 sin x cos x)²
= 3[1 + 4 sin² x cos² x – 4 sin x cos x]
= 3 + 12 sin² x cos² x – 12 sin x cos x
6(sin x + cos x)²
= 6 (sin² x + cos² x + 2 sin x cos x)
= 6 + 12 sin x cos x
4(sin^6⁡x+cos^6⁡x)
= 4[(sin² x)³ + (cos² x)³]
= 4[( sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x)]
= 4[1 – 3 sin² x cos² x]
= 4 – 12 sin² x cos² x
3 (sin⁡x-cos⁡x )^4+6(sin⁡x+cos⁡x )^2+4(sin^6⁡x+cos^6⁡x )
= 3 + 12 sin² x cos² x – 12 sin x cos x + 6 + 12 sin x cos x + 4 – 12 sin² x cos² x
= 13

Q4. If a certain amount of money becomes 64000 in 2 years and Rs. 125000 in 5 years. Find the principal & rate of interest being compounded yearly?
(a) 50%, 40900
(b) 25%, 40960
(c) 35%, 40970
(d) None of these
Ans.(b)
Sol. 64000 = P(1+r/100)^2 …(i)
125000 = P(1+r/100)^5 …(ii)
Dividing (i) & (ii)
125000/64000=(1+r/100)^3
125/64=(1+r/100)^3
5/4=1+r/100
(⇒ r = 25%)
64000=P(1+25/100)^2
64000=P×(5×5)/(4×4)
(P=40960)

Q5. S.I & C.I on a certain sum of money is 600 Rs. & 610 Rs. respectively. Find the principal & rate of interest if time is 2 years?
(a) 9000
(b) 8000
(c) 7000
(d) 12000
Ans.(a)
Sol.

Rate = (310-300)/300×100=10/3%
S.I=  (P×R×T)/100
600=(P×10×2)/(3×100)
P = 9000 Rs.

Q6. If the difference between S.I & C.I on a sum of Rs. 5000, for 2 years is Rs. 128. Find the rate of interest?
(a) 10%
(b) 16%
(c) 20%
(d) 25%

Q7. If x + (1/x) = 6, then the value of (x^4  + 1/x^2 )/(x^2  – 5x + 1) will be
(a) 70
(b) 188
(c) 198
(d) 220

Q8. If 3x –(1/3x) = 18, then the value of x^2+1/(81x^2 ) is ?
(a) 345/9
(b) 326/9
(c) 341/9
(d) 351/9

Q9. If 4a + 1/3a = 4, then find the value of 9a^2+1/(16a^2 ) ?
(a) 25/2
(b) 15/2
(c) 35/2
(d) 45/2

Q10. If n = 7 + 4√3, then find the value of √n+1/√n ?
(a) –4
(b) -2√3
(c) 2√3
(d) 4

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