Important Quantitative Aptitude Questions For SSC CGL 2018 : 23rd July (Solutions)

Important Quantitative Aptitude Questions For SSC CGL : 6th May 2018


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. If sin⁡θ+cos⁡θ=a and sec⁡θ+cosec⁡θ=b,  then value of  b(a^2–1) is equal to : 
यदि sin⁡θ + cos⁡θ = a और sec θ + cosec⁡θ = b, तो b (a ^ 2-1) का मान बराबर है:
(a) 2a
(b) 3a
(c) 0
(d) 2ab


Q2. If micron = 10,000 angstroms, then100 angstroms is what percent of10 Microns ?     
यदि माइक्रोन = 10,000 एंजस्ट्रॉम, तो 100 एन्स्ट्रोम्स, 10 माइक्रोन के कितने प्रतिशत है?
(a) 0.0001%
(b) 0.001%
(c) 0.01%
(d) 0.1%


Q3. If a² + b² + c² = 2 (3a – 4b – 6c) – 61. Find value of (a – b – c).  
यदि a² + b² + c² = 2 (3a – 4b – 6c) – 61 है. तो (a – b – c) का मान ज्ञात कीजिए.
(a) 13
(b) 10
(c) 7
(d) 9


Q4. Work done by (x + 4) men in (x + 5) days is equal to the work done by (x – 5) men in (x + 20) days. The value of x is 
(X + 4) पुरुषों द्वारा (x + 5) दिनों में किया गया कार्य (x + 5) पुरुषों द्वारा (x + 20) दिनों में किए गए कार्य के बराबर है. X का मान है
(a) 20
(b) 25
(c) 30
(d) 15


Q5. What is the number of distinct triangles with integral valued sides and perimeter as 14? 
अभिन्न मूल्यवान पक्षों और परिधि के साथ 14 के रूप में विशिष्ट त्रिकोणों की संख्या क्या है?
(a) 7
(b) 4
(c) 3
(d) 6


Q6. If x=√((√5+1)/(√5–1)), then the value of 5x² – 5x – 1 is  
यदि x=√((√5+1)/(√5–1)) तो 5x² – 5x – 1 का मान है
(a) 0
(b) 3
(c) 4
(d) 5


Q7. Find the area (in cm2) of given quadrilateral ABDC.
दिए गए चतुर्भुज ABDC का क्षेत्रफल (वर्गसेमी में) ज्ञात कीजिए.



(a) 30√3
(b) 36
(c) 48
(d) 36√3


Q8. The train A left Delhi at noon sharp. Four hours later, another train B started from Delhi in the same direction. The train B overtook the train A at 10 p.m. Find the average speed of the both trains over this journey if the sum of their speed is 80 km/h.
ट्रेन A ने दिल्ली को दोपहर में छोड़ती है. चार घंटे बाद, एक अन्य ट्रेन B समान दिशा में दिल्ली से चलती है. ट्रेन B, ट्रेन A को अपराहन 10 बजे पर पीछे छोड़ती है. इस यात्रा में दोनों ट्रेनों की औसत गति ज्ञातक कीजिए, यदि उनकी गति का योग 80 किमी / घंटा है
(a) 40 km/h / 40 किमी / घंटा
(b) 50 km/h / 50 किमी / घंटा
(c) 37.5 km/h / 37.5 किमी / घंटा
(d) 36km/h / 36 किमी / घंटा


Q9. The third proportional to (x/y+y/x) and (√(x^2+y^2 )) is
(x/y+y/x) और (√(x^2+y^2 )) का तीसरा अनुपात है 
(a) xy
(b) √xy
(c) ∛xy
(d) ∜xy


Q10. 1/9+[9999 (8990/8991)]×999 is equal to
1/9+[9999 (8990/8991)]×999  बराबर है 
(a) 1
(b) 9990000
(c) 9988999
(d) 9999000


Q11. A sum of Rs. 3200 amounts to Rs. 3362 at 10% p.a. compounded quarterly. Find the time period.
3200 रुपये की राशि 3362 10% प्रतिवर्ष तिमाही संयोजित पर 3362 रुपये हो जाती है. समय अवधि ज्ञात कीजिये.
(a) ¾ year / वर्ष
(b) ½ year / वर्ष
(c) 2 year / वर्ष
(d) ¾ years / वर्ष


Q12. If a secθ + b tanθ =1 and a² sec²θ – b² tan²θ = 5, then a²b² + 4a² is equal to
यदि a secθ + b tanθ = 1 और a² sec²θ – b² tan²θ = 5, तो a²b² + 4a² बराबर है 
(a) 9b²
(b) 9/a^2
(c) (–2)/b
(d) 9


Q13. If x varies inversely as (y² – 1) and is equal to 24 when y = 10, then the value of x when y = 5 is 
यदि X, (y² – 1) के रूप में विपरीत बदल जाता है और 24 के बराबर हो जाता है जब y = 10 हो, तो y = 5 होने पर x का मान होता है
(a) 99
(b) 12
(c) 34
(d) 100


Q14. Which term of the AP: 121, 117, 113……… is its first negative term.
121, 117, 113 ……… AP का कौन सा शब्द इसका पहला नकारात्मक शब्द है
(a) 31st / 31वां
(b) 32nd / 32वां
(c) 30th / 30वां
(d) 34th / 34वां


Q5. If cot θ + cos θ = p and cot θ – cos θ = q, then (p² – q²)² in terms of p and q is —  
यदि P और Q के सम्बन्ध में cot θ + cos θ = p और cot θ – cos θ = q है, तो (p² – q²) ² है –
(a) 16pq
(b) 8pq
(c) 4pq
(d) 12pq


Solutions
S1. Ans.(a)
Sol.
 b(a^2–1)=(sec⁡θ+cosec⁡θ )((sin⁡θ+cos⁡θ )^2–1)
=((sin⁡θ+cos⁡θ)/(sin⁡θ  cos⁡θ ))(1+2 sin⁡θ  cos⁡θ–1) 
 =2 (sin⁡θ+cos⁡θ )
= 2a 


S2. Ans. (d) 
Sol. 
Let 100 angstrom = x % of 10 Microns 
Then, 100 angstrom = x % of 100000 angstroms
⇒ x = (100×100)/100000=0.1 %


S.3 Ans.(a)
Sol.
a² + b² + c² = 6a – 8b – 12c – 61
or a² + 9 – 6a + b² + 16 + 8b + c² + 36 + 12c = 0 
(a – 3)² + (b + 4)² + (c + 6)² = 0
Here, a = 3, b = –4, c = –6
So, 
(a – b – c) = (3 + 4 + 6) = 13


S4. Ans.(a)
Sol.
According to the question
M₁D₁ = M₂D₂
(x + 4) (x + 5) = (x – 5) (x + 20) 
= x² + 9x + 20 = x² + 15x – 100
x = 20


S5. Ans. (b) 
Sol. 
The sum of length of two sides > the length of the third side 
So, the maximum length of any particular side can be 6 units 
Now if a = 6, then b + c = 8, so the possible sets are (6, 6, 2), (6, 5, 3) and (6, 4, 4) 
If a = 5 then b + c = 9
So possible set = 5, 5, 4
So the number of distinct triangle = 4 


S6. Ans.(c)
Sol. x=√(((√5+1))/((√5–1)))
x=√(((√5+1))/(√5–1)×((√5+1))/(√5+1))=√((√5+1)^2/(5–1)) 
x=(√5+1)/2 
Now, 5x² – 5x – 1 
=5((√5+1)/2)^2–5((√5+1)/2)–1 
=5((5+1+2√5)/4)–(5√5+5)/2–1 
=5((3+√5)/2)–(5√5+5)/2–1 
=(15+5√5–5√5–5–2)/2=8/2=4 


S7. Ans.(d)
Sol.
                
∠A & ∠D are supplementary 
Hence ∠D = 180° – 60° = 120° 
ATQ
Required Area = 1/2×15×8×sin⁡〖60°〗+1/2×6×4×sin⁡120  
 =30 √3+6√3  
= 36√3 cm2


S8. Ans.(C)
Sol. 
12 : 00noon      A → 10 p.m. 
4 : 00 pm           B → 10 p.m
Both the trains are moving in the same direction. Train B started 4 hours later from train A. 
A : B
Time 10 : 6
Speed 3 :
Let speed A= 3x and B = 5x
From question 
3x + 5x = 8x → 80 
           x→10
Speed of A = 3 × 10 = 30 km/h 
Speed of B = 5 × 10 = 50 km/h
∴Average speed=(2×30×50)/(50+30)=37.5 kmph 


S9. Ans.(a)
Sol.
Let third proportional = a
Then (√(x^2+y^2 ))^2=(x/y+y/x)×a 
a×(x^2+y^2)/xy=x^2+y^2 
a = xy 


S10. Ans.(b)
Sol.
=1/9+[9999×999+8990/8991×999]
=1/9+[(10000-1)999+8990/9]
=1/9+8990/9+(10000-1)999
=[8991/9+(10000-1)999]
=999[1+10000-1]
= 9990000


S11. Ans.(b)
Sol.
 3362=3200(1+10/400)^(4×t)
 or  1681/1600=(41/40)^4t
 or (41/40)^2=(41/40)^4t
 or ▭(t=1/2 year)


S12. Ans.(a)
Sol.
a sec⁡θ=1–b tan⁡θ …(i)
⇒a^2  sec^2⁡θ=1+b^2  tan^2⁡θ–2b tan⁡θ…(ii) 
Also

⇒1 – 2b tan⁡θ = 5 
⇒ b tan⁡θ=–2
⇒ b^2   sin^2⁡〖 θ〗/cos^2⁡θ =4
⇒ (b^2 (1-cos^2⁡θ))/cos^2⁡θ =4
⇒ b^2-b^2  cos^2⁡〖θ=4 cos^2⁡θ 〗
⇒ cos^2⁡θ=b^2/(b^2+4)   ….(iii)
Put b tan⁡θ = – 2 in (i) 
 a sec⁡θ=3
 a^2=9 cos^2⁡θ
cos^2⁡θ=a^2/9 …(iv)
By (iii) & (iv) 
(9b^2)/(b^2+4)=a^2 
a^2 (b^2+4)=9b^2 


S13. Ans.(a)
Sol.
Given – x∝1/(y^2–1)
or x=k/(y^2–1)
If y = 10 then x = 24
So,
24=k/99 
k = 99 × 24 
Now, when y = 5 
Then x=(99×24)/24 
x = 99


S14. Ans.(b)
Sol.
Here a = 121 and d = -4
The first negative term will be less than zero. Moreover each term gives 1 as remainder when divided by 4. So the last positive term should be 5 let us check the value of n for last term as 5
5 = a + (n – 1)d
5 = 121 + (n-1)-4
-116 = -4n + 4
N = 30
Thus, a31 = 5 – 4 = 1
a32 = 1 – 4 = -3
Hence, 32nd term will be the first negative term.


S15. Ans.(a)
Sol.
p = cot θ + cos θ
q = cot θ – cos θ
p + q = 2 cot θ,  p – q = 2 cos θ 
(p² – q²) = 4 cot θ.cos θ 
(p² – q²)² = 16 cot² θ.cos² θ =16×cos^4⁡θ/sin^2⁡θ 
pq = cot² θ – cos² θ 
=cos^2⁡θ ((1–sin^2⁡θ)/sin^2⁡θ )=cos^4⁡θ/sin^2⁡θ  
 ∴(p^2–q^2 )^2=16pq 

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