The average temperature of Delhi in the first four days of June 2017 was 52 degrees. The average for the second, third, fourth and fifth days was 55 degrees. If the temperatures of the first and fifth days were in the ratio 5 : 7, then what was the temperature of Delhi on 5th June 2017?
Sum of temperature on 1st, 2nd, 3rd, and 4th days = (52 × 4) = 208----------(I)
Sum of temperature on 2nd, 3rd, 4th, and 5th days = (55 × 4) = 220---------(II)
Subtracting equation (I) from (II)
Temperature on 5th June 2017 – temperature on 1st June 2017 = 12 degrees.
Let the temperature on 1st and 5th June 2017 be 5x and 7x degrees respectively.
Then, 7x – 5x = 12
2x = 12
X = 6
Since temperature on the 5th June 2017 in Delhi = 7x = 7 × 6 = 42 degrees.
Hence, option (D) is correct.
1
जून 2017 के पहले चार दिनों में दिल्ली का औसत तापमान 52 डिग्री था. दूसरे, तीसरे, चौथे और पांचवें दिन का औसत 55 डिग्री रहा। यदि पहले और पांचवें दिन का तापमान 5:7 के अनुपात में था, तो 5 जून 2017 को दिल्ली का तापमान कितना था?
Sum of temperature on 1st, 2nd, 3rd, and 4th days = (52 × 4) = 208----------(I)
Sum of temperature on 2nd, 3rd, 4th, and 5th days = (55 × 4) = 220---------(II)
Subtracting equation (I) from (II)
Temperature on 5th June 2017 – temperature on 1st June 2017 = 12 degrees.
Let the temperature on 1st and 5th June 2017 be 5x and 7x degrees respectively.
Then, 7x – 5x = 12
2x = 12
X = 6
Since temperature on the 5th June 2017 in Delhi = 7x = 7 × 6 = 42 degrees.
Hence, option (D) is correct.
D
2
The average weekly salary per head of all employees (supervisors and labourers) is Rs. 100. The average weekly salary per head of all the supervisors is Rs. 600 while the average weekly salary per head of all the labourers is Rs. 75. Find the number of supervisors in the factory if there are 840 labourers in it.
सभी कर्मचारियों (पर्यवेक्षकों और मजदूरों) का प्रति व्यक्ति औसत साप्ताहिक वेतन 100 रुपये है। सभी पर्यवेक्षकों का प्रति व्यक्ति औसत साप्ताहिक वेतन 600 रुपये है जबकि सभी मजदूरों का प्रति व्यक्ति औसत साप्ताहिक वेतन 75 रुपये है। यदि कारखाने में 840 मजदूर हैं तो उसमें पर्यवेक्षकों की संख्या ज्ञात कीजिए।
The average score (runs/match) of Dravid before the start of World Cup was 35. In the world cup of 6 matches, his total score was 355 runs. Find the total number of matches played by him till date, if his new average after the series is 37.5.
Let the total number of matches Dravid has played, till the start of world cup be X
So, total runs scored by Dravid till the beginning of the world cup be 35 × X
After the world cup, the new average will be
35 × X + 355
X + 6
According to question,
35 × X + 355
= 37.5
X + 6
⇒ 37.5 × X + 225 = 35 × X + 355
⇒ 2.5 × x = 130
⇒ X = 52
Hence, the total number of matches played by Dravid till date = X + 6 = 52 + 6 = 58
Therefore, option (D) is correct.
3
विश्व कप शुरू होने से पहले द्रविड़ का औसत स्कोर (रन/मैच) 35 था। 6 मैचों के विश्व कप में उनका कुल स्कोर 355 रन था. उसके द्वारा अब तक खेले गए मैचों की कुल संख्या ज्ञात कीजिए, यदि श्रृंखला के बाद उसका नया औसत 37.5 है।
Let the total number of matches Dravid has played, till the start of world cup be X
So, total runs scored by Dravid till the beginning of the world cup be 35 × X
After the world cup, the new average will be
35 × X + 355
X + 6
According to question,
35 × X + 355
= 37.5
X + 6
⇒ 37.5 × X + 225 = 35 × X + 355
⇒ 2.5 × x = 130
⇒ X = 52
Hence, the total number of matches played by Dravid till date = X + 6 = 52 + 6 = 58
Therefore, option (D) is correct.
D
4
City Pretoria is 7800 km directly to the east of city Canberra . A plane takes off from Canberra at 8:00am one day and lands at Pretoria at 3:00 am the next day. Five hours later it takes off from Pretoria and lands at 2:00 pm the same day at Canberra. There is a steady wind blowing at 25 km/hr from east to west. With respect to the wind, the plane cruises at the same uniform speed in either direction, but the speed with respect to the ground is different. What is the time difference between Canberra and Pretoria?
Let the plane speed with respect to the wind be u km/hr
The wind speed is 25 km/hr
Also, let Pretoria be t hours ahead of Canberra
Now,
7800
= 19 – t ………. (i)
u – 25
7800
= 6 + t ………….(ii)
u + 25
Adding (i) and (ii), we get
7800
+
7800
= 25
u – 25
u + 25
⇒ 7800(2u) = 25(u2 – 625)
⇒ 312(2u) = (u2 – 625)
⇒ u2 – 624u – 625 = 0
⇒(u – 625) (u + 1) = 0
As, u > 0 then
u – 625 = 0 ⇒ u = 625
Substituting in (i), we get
7800
= 19 – t ⇒ 13 = 19 – t ⇒ t = 6
600
Thus, is the time difference between Canberra and Pretoria is 6 hours.
Hence, option (D) is correct.
4
सिटी प्रिटोरिया कैनबरा शहर के पूर्व में सीधे 7800 किमी दूर है। एक विमान कैनबरा से एक दिन सुबह 8:00 बजे उड़ान भरता है और अगले दिन सुबह 3:00 बजे प्रिटोरिया पहुँचता है। पांच घंटे बाद यह प्रिटोरिया से उड़ान भरता है और उसी दिन दोपहर 2:00 बजे कैनबरा पहुँचता है। पूर्व से पश्चिम की ओर 25 किमी प्रति घंटे की रफ्तार से हवा चल रही है। हवा के संबंध में, विमान किसी भी दिशा में समान गति से परिभ्रमण करता है, लेकिन जमीन के संबंध में गति अलग होती है। कैनबरा और प्रिटोरिया के बीच समय का अंतर क्या है?
Let the plane speed with respect to the wind be u km/hr
The wind speed is 25 km/hr
Also, let Pretoria be t hours ahead of Canberra
Now,
7800
= 19 – t ………. (i)
u – 25
7800
= 6 + t ………….(ii)
u + 25
Adding (i) and (ii), we get
7800
+
7800
= 25
u – 25
u + 25
⇒ 7800(2u) = 25(u2 – 625)
⇒ 312(2u) = (u2 – 625)
⇒ u2 – 624u – 625 = 0
⇒(u – 625) (u + 1) = 0
As, u > 0 then
u – 625 = 0 ⇒ u = 625
Substituting in (i), we get
7800
= 19 – t ⇒ 13 = 19 – t ⇒ t = 6
600
Thus, is the time difference between Canberra and Pretoria is 6 hours.
Hence, option (D) is correct.
D
5
In a school, there are five friends- A, B, C, D and E. The weight of A is equal to the average weight of B, C and D and the weight of B is equal to the average weight of A, C and D. The average weight of A and C is equal to the average weight of C and D. E is 30 kg heavier than C and the average weight of B and D is 60 kg. What is the average weight (in kgs) of A, B, C, D and E?
Let the weight of four friends A, B, C, D and E are a, b, c, d and e kgs respectively
From the question,
3a = b + c + d ----------- (i)
3b = a + c + d ----------(ii)
Subtracting equation (i) and (ii)
3a – 3b = b – a
b = a ---- (iii)
Adding equation (i) and (ii)
3a + 3b = a + b + 2c + 2d
a + b = c + d ---- (iv)
The average weight of A and C is equal to the average weight of C and D
It means, a + c = c + d
a = d
Put a = d in equation (iv)
We get b = c
It means, b = a = c = d
The average weight of B and D is 60 kgs
b + d = 60 × 2 = 120 kgs
So, b = d = 60 kgs = a = c
E is 30 kg heavier than C
e = (60 + 30) kgs
By solving, e = 90 kg
The sum of the weight of A, B, C, D and E = 60 + 60 + 60 + 60 + 90 = 330 kgs
The average =
330
= 66 kgs
5
Hence, option D is correct.
5
एक विद्यालय में पाँच मित्र हैं- A, B, C, D और E। A का वजन B ,C और D के औसत वजन के बराबर है और B का वजन A , C और D के औसत वजन के बराबर है। A और C का औसत भार, C और D के औसत भार के बराबर है। E, C से 30 किग्रा भारी है और B और D का औसत भार 60 किग्रा है। A, B, C, D और E का औसत वजन (किलोग्राम में) कितना है?
