Q2. दो संख्याओं का ल.स उनके म.स के 40गुना है. उनके ल.स और म.स का योग 1,476 है. यदि उनमें से एक संख्या 288 है, तो दूसरी संख्या ज्ञात कीजिये?

(a) 169

(b) 180

(c) 240

(d) 260

Q3. चार घंटियाँ क्रमश: 14, 21 और 42 मिनट के अंतराल पर बजती हैं. यदि वे सभी 11:22पूर्वाहन पर एकसाथ बजती हैं, तो वे दोबारा एकसाथ कब बजेंगी?

(a) 11:56 पूर्वाहन

(b) 12:04 अपराहन

(c) 12:06 अपराहन

(d) 11:48 पूर्वाहन

Q4. यदि एक संख्या को 15से विभाजित किया जाता है, तो शेषफल के रूप में 7 प्राप्त होता है. यदि संख्या का तिगुना 5 से विभाजित किया जाए, तो शेषफल कितना है?

(a) 5

(b) 6

(c) 7

(d) 1

Q13. A और B दो अलग अंक हैं. यदि दोनों अंकों का उपयोग करके बनाई गई दो अंकों की संख्याओं का योग एक आदर्श वर्ग है, तो (A + B) का मान कितना होगा?

(a) 9

(b) 11

(c) 13

(d) 17

Q14. एक संख्या N = 897324P64Q दोनों 8 और 9 से विभाज्य है. निम्नलिखित में से P + Q का मान क्या है?

(i). 2                (ii). 11             (iii). 9

(a) या तो (i) या (ii)

(b) या तो (ii) या (iii)

(c) या तो (i) या (ii) या (iii)

(d) इनमें से कोई नहीं

Solutions

S1. Ans.(b)

Sol.

If we look at the numbers 100 < N ≤ 105, we see only 101 and 103 do not have their factors in N (because these are primes). So, obviously the new LCM will be 101 × 103 × N. 

S2. Ans.(d)

Sol.

Assume that the total population is 100. 

30 people smoke, and out of them 6 people will be having lung cancer.

This 6 represents 80% of lung cancer patients (because they smoke).

 Hence, total percentage of population having lung cancer = 6/80×100=7.5%

S3. Ans.(c)

Sol.

■(Sugar&:&Water@1&:&4) 

Let the weight of solution be 5 units 

Weight of Sugar = 1 unit

After heating weight of solutions = 2.5 unit

So, weight of water remaining = 2.5 – wt. of Sugar 

= 1.5 unit

Ratio = 1 : 1.5

= 2 : 3 

S4. Ans.(a)

Sol.

Suppose price of one article is Re 1.

 So, price of 4 articles = Rs. 4.

 Now the whole scene can be understood as —

We are paying only Rs. 3 at the place of Rs. 4. 

So, discount = Re1. 

So, discount percentage = ¼ (discount/total price) = 25%

Also, you can use trice → (free item)/(Total item)×100

=1/4×100=25% 

S5. Ans.(a)

Ratio of Milk and Water

M :W

80:20 

1/4 th of the mixture is sold, Milk and Water both will be reduced by 25%

 

To replenish the quantity, we will add 25 litres of water. 

So, New ratio 

M :W

60:40

3:2

Water : Milk = 2 : 3 

S6. Ans.(a)

Sol.

50% of milk solution will have milk and water in the ratio

                                1 : 1

Final ratio → 10:90 

                       1   :9

Now, 

2r=30l (Volume of mixture) 

1r = 15ℓ

8r = 120l.

S7. Ans.(a)

Sol.

Expenses = F + K. V; where F is the fixed cost and V is the number of students. 

Rs. 15,000 = F + K.20 ……………(i)

Rs. 20,000 = F + K.30 ……………..(ii)

Solving (i) and (ii),

Rs. 5,000 = 10. K ⇒ K = Rs. 500

So, F = Rs. 5000

So, F + 40 K = Rs. 5,000 + 40 × 500 = Rs. 25,000

S8. Ans.(c)

Sol.

Machine P can print 100000/8=12500 copies per hour.

Machine Q can print 100000/10=10000 copies per hour.

Machine R can print 100000/12=8333 1/3 copies per hour.

All of them together can print = 30833 1/3 copies per hour.

Therefore, in 6 hours they can print = 30833.33×6=185000 copies.

Hence, option (c) is the answer. 

S9. Ans.(b)

Sol.

18! +19!= 18!(1+19)= 18! * 20

Since number of zeroes in 18! is 3 and clearly 20 has a zero, hence total number of zeroes must be 4.

 S10. Ans.(a)

Sol.

Now, 

(x-20)/30=1/3×3/2 

(x-20)/30=1/2 

x=35% 

S11. Ans.(b)

Sol.

Average speed of 1st journey = (2×75×50)/125=60 km/hr

Average speed for 2nd journey = (2×30×60)/90=40 km/hr

Average speed for culture journey = (2×60×40)/100=48 km/hr

S12. Ans.(d)

Sol.

Time taken to cover 7 km = 1 hr 12 min = 6/5 hr

So, Speed = 7/((6/5) )=35/6  km/hr

  

Ratio of time → 25: 11

Now, 36 r → 6/5

1r → 1/30 

Time taken to walk on foot = 25r 

=25/30=5/6 hr

Distance covered by foot = s × t 

=4×5/6   

=10/3 km

S13. Ans.(a)

Sol.

Let the speed of two trains be x m/s and y m/s

Case I: 

In opposite direction

Relative speed will be = (x+y) m/s

i.e., (230+190)/(x+y)=21s

420/(x+y)=21  

x+y=20 …..(i)

Case 2:  

In same direction

Relative speed =(x-y) m/s

i.e., 420/42=x-y. Hence x-y=10 ………(ii)

On solving (i) and (ii) we get x=15 m/s, y=5 m/s, 

Therefore the ratio = 3: 1

S14. Ans.(b)

Sol.

∠AOC = 160°

So, ∠ABC = 80° [Angle at the circumference]is half of the centre by the same Arc.

Also, AB = AC 

So, ∠ABC = ∠ACB = 80°

So, ∠BAC = 20°

S15. Ans.(a)

Sol.

Given, OQ = QR 

∠QRO = 15°

So, ∠QOR = 15° 

Also, 

∠OQP = 30° (Ext. angle of a ∆) 

Now, OP = OQ = r

∠OPQ = ∠OQP = 30°

So, ∠QOP = 120°.