SSC MTS Numerical Ability Practice Questions : 26th July_00.1
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SSC MTS Numerical Ability Practice Questions : 26th July

SSC MTS 2018-19 is to be conducted from 2nd August to 27th September 2019. Today, in this Quantitative Aptitude quiz we are providing quantitative aptitude practice Questions with detailed Solutions to make your practice effective for quantitative aptitude Section for SSC MTS 2018-19 Exam. Attempt this quiz and prepare yourself flawlessly. We wish you good luck for all the upcoming Exams.

Q1. There are 2 teams-A and B. If 3 people are shifted from Team A to Team B, then Team B has thrice the number of members than Team A. If 2 people are shifted from Team B to Team A, then Team B has double the number of members than Team A. How many members does Team B have originally?
2 टीम,A और B दी गयी है. यदि 3 व्यक्तियों को टीम A से टीम B में स्थानांतरित कर दिया जाता है, तो टीम B में व्यक्तियों की संख्या टीम A से तीन गुना होगी. यदि 2 व्यक्तियों को टीम B से टीम A में स्थानांतरित कर दिया जाता है तो टीम B में टीम A की तुलना में दोगुने व्यक्ति होंगे. वास्तव में टीम B में कितने सदस्य हैं?
(a) 15
(b) 18
(c) 42
(d) 45

S1. Ans.(c)
Sol. Member in team A ⇒ x
Member in team B ⇒ y
ATQ,
3 (x – 3) = y + 3
y + 3 = 3x – 9
3x – y = 12 …(i)
(y – 2) = 2(x + 2)
y – 2 = 2x + 4
2x – y = –6 …(ii)
From (i) & (ii)
x = 18
y = 42

Q2. (6² + 7² + 8² + 9² + 10² )/(√(7 + 4√3) – √(4 + 2√3) ) is equal to
(6² + 7² + 8² + 9² + 10² /(√(7 + 4√3) – √(4 + 2√3) ) किसके बराबर है
(a) 330
(b) 355
(c) 305
(d) 366

S2. Ans.(a)
Sol.
(6² +7² +8² +9² +10² )/(√(7+4√3) -√(4+2√3) )
=(36+49+64+81+100)/(√(4+3+2×2√3) -√(1+3+2√3) )
=(200+130)/(√((2+√3)² )-√((1+√3)² ))
=330/(2+√3-1-√3)=330

Q3. There are 1400 students in a school, 25% of those wear spectacles and 2/7 of those wearing spectacles are boys. How many girls in the school wear spectacles?
एक स्कूल में 1400 छात्र हैं, जिनमें से 25% चश्मा पहनते हैं और चश्मे पहनने वाले छात्रों में से से 2/7 लड़के हैं. स्कूल में कितनी लड़कियां चश्मा पहनती हैं?
(a) 250
(b) 100
(c) 200
(d) 300

S3. Ans.(a)
Sol. Student wearing spectacles=1400×1/4= 350
Boys wearing spectacles=350×2/7=100
Girls wearing spectacles= 350 – 100= 250

Q4. If x² + 1/x² = 98 (x > 0), then the value of x³ + 1/x³ is
यदि x² + 1/x² = 98 (x > 0) है,तो x³ + 1/x³ का मान ज्ञात कीजिये
(a) 970
(b) 1030
(c) –970
(d) –1030

S4. Ans.(a)
Sol.
x² + 1/x² =98
Adding 2 both sides
x² + 1/x² +2=98+2
(x+1/x)²=100
x+1/x=10
Cubing both sides
(x+1/x)³ =(10)³
x³ +1/x³ +3(x+1/x)=1000
x³ +1/x³ =1000-30= 970

Q5. The marked price of an article is Rs. 5000 but due to festive offer a certain percent of discount is declared. Mr. X availed this opportunity and bought the article at reduced price. He then sold it at Rs. 5000 and thereby made a profit of 111/9%. The percentage of discount allowed was?
एक वस्तु का अंकित मूल्य 5000 रु है लेकिन एक उत्सव के कारण एक निश्चित छूट की घोषणा की जाती है. श्रीमान X इस छुट का फायदा उठाते हैं और कम किये गए मूल्य पर वस्तु खरीदते हैं. फिर वह उसे 5000 रु में बेच देते हैं और इससे वे 111/9% का लाभ अर्जित करते हैं. दी गयी छुट का प्रतिशत कितना था?
(a) 10
(b) 31/3
(c) 71/2
(d) 111/9

S5. Ans.(a)
Sol. S.P for x ⇒ 5000
Profit % = 111/9%
=100/9%
=1/9 (Profit/C.P.)
S.P = 9 + 1 = 10
10r → 5000
1r → 500
9r → 4500 Rs.
Discount % = 500/5000 × 100
= 10%

Q6. Arun buys one kilogram of apples for Rs. 120 and sells it to Swati gaining 25%. Swati sells it to Divya and Divya again sells it for Rs. 198, making a profit of 10%. What is the profit percentage made by swati?
अरुण 120 रु में एक किलोग्राम सेब खरीदता है और 25% के लाभ पर इसे स्वाति को बेच देता है. स्वाति इसे दिव्या को बेच देती है और दिव्या फिर इसे 198 रु में बेच देती है और 10% का लाभ अर्जित करती है. स्वाति द्वारा अर्जित किया गया लाभ प्रतिशत कितना है?
(a) 25%
(b) 20%
(c) 16.67%
(d) 15%

S6. Ans.(b)
Sol. C.P of Arun = 120
S.P of Arun & C.P of Swati
=120×125/100
= 150 Rs.
S.P of Divya = 198
C.P of Divya and S.P of Swati
=198×100/110=180
Profit of Swati
=(180-150)/150×100
=30/150×100
= 20%

Q7. If 7 sin α = 24 cos α; 0 < α < π/2, then the value of 14 tan α – 75 cos α – 7 sec α is equal to
यदि 7 sin α = 24 cos α; 0 < α < π/2 है,तो 14 tan α – 75 cos α – 7 sec α का मान किसके बराबर होगा:
(a) 3
(b) 4
(c) 1
(d) 2

S7. Ans.(d)
Sol. 7 sin α = 24 cos α
tan⁡=24/7(Perpendicular/Base)
H = √(576+49)
=√625
= 25
cos⁡α=B/H=7/25
sec⁡α=1/cos⁡α =25/7
14 tan α – 75 cos α – 7 sec α
=14×24/7-75×7/25-7×25/7
= 48 – 21 – 25
= 48 – 46
= 2

Q8. A tower is observed from a point on the horizontal through the foot of the tower. The distance of this point from the foot of the tower is equal to the height of the tower. The angle of elevation of the top of the tower is
एक टावर को उसके आधार के क्षैतिज बिंदु से देखा जाता है. टावर के आधार से इस बिंदु की दूरी टॉवर की ऊंचाई के बराबर है. टॉवर का उन्न्यन्न कोण कितना है?
(a) 60°
(b) 45°
(c) 40°
(d) 30°

S8. Ans.(b)
Sol.

SSC MTS Numerical Ability Practice Questions : 26th July_50.1

tan θ = h/h
θ = 45°

Q9. At present ages of a father and son are in the ratio of 7 : 3 and they will be in the ratio 2 : 1 after 10 years. What is the present age of father?
वर्तमान में पिता और पुत्र की आयु का अनुपात 7: 3 है और 10 वर्षों के बाद यह अनुपात 2:1 होगा. पिता की वर्तमान आयु कितनी है?
(a) 70 years/वर्ष
(b) 65 years/वर्ष
(c) 60 years/वर्ष
(d) 50 years/वर्ष

S9. Ans.(a)
Sol.

SSC MTS Numerical Ability Practice Questions : 26th July_60.1

(8 – 7) r = 10
1r = 10
Present Age of father = 7 × 10
= 70 years

Q10. John cycling at a constant speed of 10 km/hr., reaches his school in time. If he cycles at a constant speed of 15 km/hr., he reaches his school 12 minutes early. Number of km he has to cycle for his school is :
जॉन 10 किमी/घंटा की स्थिर गति से साइकिल चलाते हुए, समय पर अपने स्कूल तक पहुंचता है. यदि वह 15 किमी/घंटा की स्थिर गति से साइकिल चलाता, तो वह 12 मिनट पहले अपने स्कूल तक पहुंच जाता है. उसे अपने स्कूल तक जाने के लिए कितनी किमी तक साइकिल चालानी है?
(a) 4
(b) 6
(c) 9
(d) 12

S10. Ans.(b)
Sol.
Ratio of speed ⇒10:15⇒2:3
Ratio of time ⇒3 : 2
(3 – 2)r ⇒ 12 minutes
3r → 36 minutes
Distance = 10 km/hr × 36/60
= 6 km

SSC MTS Numerical Ability Practice Questions : 26th July_70.1

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