Question 5
एक दुकानदार अपनी दुकान में बचे चावलों को दो हिस्सो
में बांटता है जोकि 5:2 है। यदि बड़ा हिस्से के चावलों
का वजन 35 किलो है तो छोटा हिस्सा कितने किलो का
BITI ?A shopkeeper divides the remaining rice
of his shop in two parts which is 5:2. If the
larger portion weighs 35 kg, how much will the
smaller portion weigh?
Answers
Answer:
The smaller portion weighs 14 kg.
Step-by-step explanation:
Given that:
A shopkeeper divides the remaining rice of his shop in two parts which is 5 : 2.
The larger portion weighs 35 kg.
To Find:
How much will the smaller portion weigh?
Let us assume:
The smaller portion weighs x kg.
Finding the weight of smaller portion:
According to the question.
⇒ 35 : x = 5 : 2
⇒ 35/x = 5/2
Cross multiplication.
⇒ 5 × x = 35 × 2
⇒ 5x = 70
⇒ x = 70/5
⇒ x = 14
∴ The weight of smaller portion = 14 kg
Question:-
Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of the next four terms.
To Find:-
Find the common difference.
Solution:-
Formula to be Used:-
\sf \: T_n = a + ( n - 1 )dT
n
=a+(n−1)d
We can also write as:-
\longrightarrow\sf \: a_1 + a_2 + a_3 + a_4 = \dfrac { 1 } { 2 } ( a_5 + a_6 + a_7 + a_8 )⟶a
1
+a
2
+a
3
+a
4
=
2
1
(a
5
+a
6
+a
7
+a
8
)
\longrightarrow\sf \: a + ( a + d ) + ( a + 2d ) + ( a + 3d ) = \dfrac { 1 } { 2 } ( a + 4d + a + 5d + a + 6d + a + 7d )⟶a+(a+d)+(a+2d)+(a+3d)=
2
1
(a+4d+a+5d+a+6d+a+7d)
\longrightarrow\sf \: 4a + 6d = \dfrac { 1 } { 2 } ( 4a + 22d )⟶4a+6d=
2
1
(4a+22d)
\longrightarrow\sf \: 4( 5 ) + 6d = \dfrac { 1 } { 2 } \times 2 \times ( 2( 5 ) + 11d )⟶4(5)+6d=
2
1
×2×(2(5)+11d)
\longrightarrow\sf \: 20 + 6d = 10 + 11d⟶20+6d=10+11d
\longrightarrow\sf \: 20 - 10 = 11d - 6d⟶20−10=11d−6d
\longrightarrow\sf \: 5d = 10⟶5d=10
\longrightarrow\sf \: d = \cancel\dfrac { 10 } { 5 }⟶d=
5
10
\longrightarrow\sf \: d = 2⟶d=2
Hence ,
Common difference is 2
Question:-
Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of the next four terms.
To Find:-
Find the common difference.
Solution:-
Formula to be Used:-
\sf \: T_n = a + ( n - 1 )dT
n
=a+(n−1)d
We can also write as:-
\longrightarrow\sf \: a_1 + a_2 + a_3 + a_4 = \dfrac { 1 } { 2 } ( a_5 + a_6 + a_7 + a_8 )⟶a
1
+a
2
+a
3
+a
4
=
2
1
(a
5
+a
6
+a
7
+a
8
)
\longrightarrow\sf \: a + ( a + d ) + ( a + 2d ) + ( a + 3d ) = \dfrac { 1 } { 2 } ( a + 4d + a + 5d + a + 6d + a + 7d )⟶a+(a+d)+(a+2d)+(a+3d)=
2
1
(a+4d+a+5d+a+6d+a+7d)
\longrightarrow\sf \: 4a + 6d = \dfrac { 1 } { 2 } ( 4a + 22d )⟶4a+6d=
2
1
(4a+22d)
\longrightarrow\sf \: 4( 5 ) + 6d = \dfrac { 1 } { 2 } \times 2 \times ( 2( 5 ) + 11d )⟶4(5)+6d=
2
1
×2×(2(5)+11d)
\longrightarrow\sf \: 20 + 6d = 10 + 11d⟶20+6d=10+11d
\longrightarrow\sf \: 20 - 10 = 11d - 6d⟶20−10=11d−6d
\longrightarrow\sf \: 5d = 10⟶5d=10
\longrightarrow\sf \: d = \cancel\dfrac { 10 } { 5 }⟶d=
5
10
\longrightarrow\sf \: d = 2⟶d=2
Hence ,
Common difference is 2
Question:-
Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of the next four terms.
To Find:-
Find the common difference.
Solution:-
Formula to be Used:-
\sf \: T_n = a + ( n - 1 )dT
n
=a+(n−1)d
We can also write as:-
\longrightarrow\sf \: a_1 + a_2 + a_3 + a_4 = \dfrac { 1 } { 2 } ( a_5 + a_6 + a_7 + a_8 )⟶a
1
+a
2
+a
3
+a
4
=
2
1
(a
5
+a
6
+a
7
+a
8
)
\longrightarrow\sf \: a + ( a + d ) + ( a + 2d ) + ( a + 3d ) = \dfrac { 1 } { 2 } ( a + 4d + a + 5d + a + 6d + a + 7d )⟶a+(a+d)+(a+2d)+(a+3d)=
2
1
(a+4d+a+5d+a+6d+a+7d)
\longrightarrow\sf \: 4a + 6d = \dfrac { 1 } { 2 } ( 4a + 22d )⟶4a+6d=
2
1
(4a+22d)
\longrightarrow\sf \: 4( 5 ) + 6d = \dfrac { 1 } { 2 } \times 2 \times ( 2( 5 ) + 11d )⟶4(5)+6d=
2
1
×2×(2(5)+11d)
\longrightarrow\sf \: 20 + 6d = 10 + 11d⟶20+6d=10+11d
\longrightarrow\sf \: 20 - 10 = 11d - 6d⟶20−10=11d−6d
\longrightarrow\sf \: 5d = 10⟶5d=10
\longrightarrow\sf \: d = \cancel\dfrac { 10 } { 5 }⟶d=
5
10
\longrightarrow\sf \: d = 2⟶d=2
Hence ,
Common difference is 2
Question:-
Find the common difference of an A.P. whose first term is 5 & sum of the first four terms is half the sum of the next four terms.
To Find:-
Find the common difference.
Solution:-
Formula to be Used:-
\sf \: T_n = a + ( n - 1 )dT
n
=a+(n−1)d
We can also write as:-
\longrightarrow\sf \: a_1 + a_2 + a_3 + a_4 = \dfrac { 1 } { 2 } ( a_5 + a_6 + a_7 + a_8 )⟶a
1
+a
2
+a
3
+a
4
=
2
1
(a
5
+a
6
+a
7
+a
8
)
\longrightarrow\sf \: a + ( a + d ) + ( a + 2d ) + ( a + 3d ) = \dfrac { 1 } { 2 } ( a + 4d + a + 5d + a + 6d + a + 7d )⟶a+(a+d)+(a+2d)+(a+3d)=
2
1
(a+4d+a+5d+a+6d+a+7d)
\longrightarrow\sf \: 4a + 6d = \dfrac { 1 } { 2 } ( 4a + 22d )⟶4a+6d=
2
1
(4a+22d)
\longrightarrow\sf \: 4( 5 ) + 6d = \dfrac { 1 } { 2 } \times 2 \times ( 2( 5 ) + 11d )⟶4(5)+6d=
2
1
×2×(2(5)+11d)
\longrightarrow\sf \: 20 + 6d = 10 + 11d⟶20+6d=10+11d
\longrightarrow\sf \: 20 - 10 = 11d - 6d⟶20−10=11d−6d
\longrightarrow\sf \: 5d = 10⟶5d=10
\longrightarrow\sf \: d = \cancel\dfrac { 10 } { 5 }⟶d=
5
10
\longrightarrow\sf \: d = 2⟶d=2
Hence ,
Common difference is 2