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Algebra Level 1 Test 10
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Question 1 |
Sonu is 4 yr younger than Manu while dolly is 4 yr younger than Sumit but 1/5 times as old as Sonu. If Sumit is eight years old, how many times as old is Manu as Dolly?
6 | |
1/2 | |
3 | |
None of these |
Question 1 Explanation:
From the given conditions we come to know that
Sonu’s age = Manu’age - 4
Dolly’s age = Sumit’s age - 4
Therefore Dolly‘s age = 1/5 sonu
Therefore the age of Sumit = 8 yr,
Dolly = 4 yr,
The age of Sonu = 20 yr
and Manu = 24 yr
Therefore we can say that Manu = 6 x Dolly
Sonu’s age = Manu’age - 4
Dolly’s age = Sumit’s age - 4
Therefore Dolly‘s age = 1/5 sonu
Therefore the age of Sumit = 8 yr,
Dolly = 4 yr,
The age of Sonu = 20 yr
and Manu = 24 yr
Therefore we can say that Manu = 6 x Dolly
Question 2 |
A group consigning of 25 teachers, 20 engineers, 18 doctors and 12 salesmen visited a fair spent Rs. 1330 altogether. It was found that 5 teachers spent as much as 4 engineers; 12 engineers spent as much as 9 doctors and 6 doctors spent as much as 8 salesmen. If every person in a professional group spent the same amount, the amount spent by each engineer is
Rs. 18 | |
Rs.17.50 | |
Rs. 14 | |
Rs. 21 |
Question 2 Explanation:
Let we make the equations for the given conditions
25T + 20E + 18D + 12S = Rs. 1330 ………………..1
Also Given 5 teachers spent as much as 4 engineers,
Therefore 5T = 4E
= T = 4/5E
Also 12E = 9D
D= 12/9E
Therefore 6D =8S
Hence S=6D/8 = 6/8 x 12/9E = E
Therefore From Eq. (i),
25 x 4/5E +20 E + 18 x 12/9E + 12E = 1330
76E = 1330
E =Rs. 17.50
25T + 20E + 18D + 12S = Rs. 1330 ………………..1
Also Given 5 teachers spent as much as 4 engineers,
Therefore 5T = 4E
= T = 4/5E
Also 12E = 9D
D= 12/9E
Therefore 6D =8S
Hence S=6D/8 = 6/8 x 12/9E = E
Therefore From Eq. (i),
25 x 4/5E +20 E + 18 x 12/9E + 12E = 1330
76E = 1330
E =Rs. 17.50
Question 3 |
In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in the family?
4 | |
3 | |
2 | |
5 |
Question 3 Explanation:
Let number of sister each son have = 2s
Let number of brothers each son have = s
Therefore the number of sons in the family = (s+1)
Daughters = 2s
Now from the question (2s -1) = (s-1)
Therefore s = 2
Therefore number of sons in the family = 3
Let number of brothers each son have = s
Therefore the number of sons in the family = (s+1)
Daughters = 2s
Now from the question (2s -1) = (s-1)
Therefore s = 2
Therefore number of sons in the family = 3
Question 4 |
The sum of ages of a man and his son is 100 years now, five years back their ages were in the ratio of 2:1. What will be the ratio of their ages after 10 years?
3:4 | |
5:3 | |
4:5 | |
5:6 |
Question 4 Explanation:
Let the age of man = p years
Therefore the age of his son would be = 100- p
Therefore {(p – 5)/(100 – p)-5}= 2/1
On solving for p we gat
p = 65
Therefore the ratio of their ages would be after 10 years
= {(p +10)/(100-p)+10}
= (p +10)/(110 –p)= 5:3
Therefore the age of his son would be = 100- p
Therefore {(p – 5)/(100 – p)-5}= 2/1
On solving for p we gat
p = 65
Therefore the ratio of their ages would be after 10 years
= {(p +10)/(100-p)+10}
= (p +10)/(110 –p)= 5:3
Question 5 |
The product of Richa’s age 5 year ago with her age 9 years hence is 15. Richa’s Present age is
6 | |
8 | |
49 | |
10 |
Question 5 Explanation:
Let the present age of Richa = p yrs
Therefore
(p -5)(p +9) = 15
Hence there will be two values for p
One is -10 and other is 6
So the right value will be 6
Therefore
(p -5)(p +9) = 15
Hence there will be two values for p
One is -10 and other is 6
So the right value will be 6
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